Imported and Manufactured Food Program Inspection Manual
Chapter 10: Calculations

This page is part of the Guidance Document Repository (GDR).

Looking for related documents?
Search for related documents in the Guidance Document Repository

10.0 Scope

This chapter provides examples of calculations that inspectors may need to perform while conducting inspections and complaint investigations. In addition to a handy conversion chart, examples of calculations used in the following situations are included:

  • addition of chemicals to achieve concentrations in parts per million (ppm)
  • addition of vitamins A and D to fluid milk
  • degree-hours for fermentation of ready-to-eat meats
  • calculations for common food additives (nitrate, nitrite, sulphur dioxide)
  • calculations for allergens in finished goods

10.1 Metric/Imperial Conversion Tables

Table 1: Length
Starting UnitFormulaDestination Unit
inches (in) Multiply by 2.54 centimetres (cm)
centimetres (cm) Divide by 2.54 inches (in)
feet (ft) Multiply by 0.305 metres (m)
metres (m) Divide by 0.305 feet (ft)
yards (yds) Multiply by 0.914 metres (m)
metres (m) Divide by 0.914 yards (yds)
mile (mi) Multiply by 1.609 kilometres (km)
kilometres (km) Divide by 1.609 mile (mi)
Table 2: Area
Starting UnitFormulaDestination Unit
square inches (sq in) Multiply by 6.452 square centimetres (cm2)
square centimetres (cm2) Divide by 6.452 square inches (sq in)
square feet (sq ft) Multiply by 0.092 square metres (m2)
square metres (m2) Divide by 0.092 square feet (sq ft)
Table 3: Volume
Starting UnitFormulaDestination Unit
cubic inches (cu in) Multiply by 16.388 cubic centimetres (cm3)
cubic centimetres (cm3) Divide by 16.388 cubic inches (cu in)
Imperial fluid ounce (imp fl oz) Multiply by 28.413 millilitres (mL)
millilitres (mL) Divide by 28.413 Imperial fluid ounce (imp fl oz)
US fluid ounce (US fl oz) Multiply by 29.66 millilitres (mL)
millilitres (mL) Divide by 29.66 US fluid ounce (US fl oz)
Imperial (imp gal) Multiply by 4.546 litres (L)
litres (L) Divide by 4.546 Imperial (imp gal)
US gallon (US gal) Multiply by 3.785 litres (L)
litres (L) Divide by 3.785 US gallon (US gal)
Table 4: Weight
Starting UnitFormulaDestination Unit
ounces (oz) Multiply by 28.38 grams (g)
grams (g) Divide by 28.38 ounces (oz)
pounds (lbs) Multiply by 0.453 kilograms (kg)
kilograms (kg) Divide by 0.453 pounds (lbs)
tonne (t) Multiply by 1000 kilograms (kg)
kilograms (kg) Divide by 1000 tonne (t)
(metric ton)
short ton Multiply by 907.185 kilograms (kg)
kilograms (kg) Divide by 907.185 short ton
foot-pounds Multiply by 1.356 Newton-metres (N-m)
Newton-metres (N-m) Divide by 1.356 foot-pounds
Table 5: Temperature
Starting UnitFormulaDestination Unit
°Fahrenheit F - 32) × 5/9 °Celsius
°Celsius C × 9/5) + 32 °Fahrenheit

10.2 Concentration in parts per million (ppm) of Substances in Water

It may be easier to understand the concept of parts per million (ppm) if we consider that one ppm is one part out of one million parts. For example, one gram in one million grams or one millilitre out of one million millilitres:

1 g / 1,000,000 g
or
1 mL / 1,000,000 mL
or
1 part / 1,000,000 parts

It is also commonly understood that one ppm is equal to 1 mg of solute in 1 L of solution:

1 mg solute/1,000 mL solution
or
1 mg solute/1 L solution
or
0.001 g solute/1 L solution
or
0.000001 kg solute/1 L solution

For the sake of comparison:

  • 1 mg solute/L solution = 1 ppm
  • 4.546 mg solute/imp gal solution = 1 ppm
  • 3.785 mg solute/US gal solution = 1 ppm

Note: assume that 1 millilitre of water is equal to 1 gram of water (1 mL H2O = 1 g H2O).

To determine how much of a chemical must be added to water to reach a desired concentration in parts per million, the following information is required:

  • Desired concentration of final solution
  • Final volume of solution
  • Concentration or amount of active ingredient/chemical in substance/chemical being added to the water

Once you have determined the required information, use the following formula:

  • Amount of solute required = [Desired final concentrations of active ingredient in ppm (mg active ingredient/L) × Total volume of tank (L final volume)] / Concentration of solute in ppm

10.2.1 Diluting sodium hypochlorite (bleach) solutions

Example:

Gravelle's Veggies-to-Go wants to sanitize their inspection tables in water containing 200 ppm of chlorine (Cl). How much standard bleach should be added to his 1000 litre sanitizer tank?

What we know:

  • Final volume of solution = 1000 L
  • Desired ppm of final solution = 200 ppm

We will need to determine the amount of chlorine in standard bleach. Commercial bleach is a solution of sodium hypochlorite (NaClO) typically sold at a concentration of 5.25% 'available chlorine'. It is also available at higher concentrations (e.g. 12%) so it is important to verify the concentration on the label. For this example, we will assume that the bleach is 5.25% available chlorine (typical household bleach).

5.25% available chlorine = 5.25g NaClO/100mL = 52,500mg NaClO/L or ppm

We now know there is 52,500 ppm of available chlorine in the solute (bleach). We can now solve for the amount of solute required to have 200 ppm of available chlorine in a 1000 L tank.

Given that the ratio between the desired concentration (200 ppm) and the concentration of chlorine in the bleach (52,500 ppm) must be the same as the ratio between the final volume (1000L) and the amount of bleach required:

  • C1 × V1 = C2 × V2
  • V2 = (C1 × V1) / C2
  • V2 = (200 ppm × 1000 L) / 52, 500 ppm
  • V2 = 3.81 L (or 4 L)

Therefore, Gravelle's Veggies-to-Go should add 4 L of standard bleach to the 1000 L tank in order to obtain 200 ppm of chlorine in the tank.

The availability of chlorine in solution to kill microorganisms is highly dependent on the pH of the water, and the amount of dissolved minerals and organic matter. It is highly recommended that manufacturers verify the concentration of their chlorine solutions using test strips, or other methods, and adjust the concentration accordingly. The above calculation should be used to determine a starting point only.

10.3 Vitamins A and D in Fluid Milk

Division 8 of the Food and Drug Regulations requires that fluid milk contain enough added vitamin D that a reasonable daily intake of the milk contains between 300 and 400 International Units (IU) of vitamin D. Skimmed and partially skimmed milk must also contain enough added vitamin A that a reasonable daily intake of the milk contains between 1200 IU and 2500 IU of vitamin A. As well, evaporated and evaporated skimmed milk must contain enough vitamin C that a reasonable daily intake of the milk contains between 60 and 75 milligrams of vitamin C.

Health Canada has determined that a reasonable daily intake of milk is 852mL (Schedule K: Reasonable Daily Intake for Various Foods, Food and Drug Regulations), thus allowing the calculation of the desired vitamin concentration ranges and target levels (Table 6):

Table 6 - Target Fortification Levels for Fluid and Evaporated Milk
VitaminRangeTarget Level
A 140.8 – 293 IU / 100 mL 216.9 IU / 100 mL
D 35.2 – 46.9 IU / 100 mL 41.0 IU /100 mL
C 14.08 – 17.6 mg / 100 mL 15.84 mg / 100 mL

The appropriate vitamin fortification target level is the midpoint of the concentration range.

10.3.1 Calculating Vitamin Concentration in Fluid Milk

There are two basic vitamin concentration calculations typically used in the fluid milk industry:

  1. Theoretical Value, used by dairy personnel to determine how much vitamin premix should be added to achieve the desired vitamin concentration in the product. The pasteurization method used (batch or continuous flow) will affect how this value is calculated, as will the use of diluted vitamin premix
  2. Calculated Value, used to estimate the actual vitamin concentration in the milk. The use of diluted vitamin premix will also affect the calculation of this value

The following calculation information is taken from the Dairy Products Inspection Manual.

10.3.1.1 Determining the Concentration of Diluted Vitamin Solution

If the vitamin solution has been diluted, the concentration of the diluted vitamin solution must be determined prior to calculating either the Theoretical Value or the Calculated Value.

The following information is required:

  • Concentration of the vitamin premix
  • Volume of the vitamin premix used
  • Volume of the diluted vitamin solution to be prepared

Once this information has been obtained, use the following formula:

Cd × Vd = Cp × Vp

where Cd = concentration of diluted vitamin solution

Example:

The concentration of a vitamin D premix is 205,000 IU/mL. 24 mL of this premix is used in the preparation of 500 mL of diluted vitamin solution. What is the concentration of the diluted vitamin solution?

Using

  • Cd = (Cp × Vp) / Vd
  • Cd (IU/mL) = (205,000 IU/mL × 24 mL) / 500 mL
  • Cd = 9840 IU/mL

This value will be used for "C" when calculating the Theoretical Value or the Calculated Value in the next two sections.

10.3.1.2 Calculating Theoretical Values for Batch Method Pasteurization

Information required:

  • T = Theoretical Concentration of Vitamin in the Final Product (IU/100 mL)
  • V = Volume of Vitamin Premix used (mL)
  • C = Concentration of Vitamin Premix (IU/mL)
  • Q = Quantity of Milk to be Fortified (mL)Footnote 7

Equation used:

T = VC/Q

Example:

  • T = Theoretical Concentration of Vitamin in the Final Product (IU/100mL)
  • V = 3.35mL
  • C = 205,000 IU/mL (of Vitamin D)
  • Q = 1800L (1800L × 1000mL/L = 1800000mL)

T = (205,000 IU/mL × 3.35mL) / 1,800,000mL

T = 0.382 IU/mL of vitamin D

Multiply by 100 to convert T to IU/100mL

Therefore:

(0.3815 IU/mL) (100) = 38.15 IU/100mL

T = 38.15 IU/100mL of Vitamin D

10.3.1.3 Calculating Theoretical Values for Continuous Flow Method Pasteurization

Information required:

  • T = Theoretical Concentration of Vitamin in the Final Product (IU/100mL)
  • C = Concentration of Vitamin Premix (or concentration of diluted Vitamin solution) (IU/mL)
  • P = Vitamin Pump Speed (mL/minute)
  • F = Flow rate of Milk (mL/minute)Footnote 8

Equation used:

T = CP/F

Example 1:

  • T = Theoretical Concentration of Vitamin in the Final Product (IU/100mL)
  • C = 95,000 IU/mL (of Vitamin A)
  • P = 3.5mL/minute
  • F = 292L/minute (292L/min × 1000mL/L = 292,000mL/min)
  • T = CP/F
    T = (95,000 IU/mL × 3.5mL/minute) / 292mL/minute
  • T = (332,500 IU/minute) / (292,000mL/minute)
  • T = 1.14 IU/mL of Vitamin A

This needs to be converted to IU/100mL. Do this by multiplying your calculated T value by 100.

Therefore:

  • (1.14 IU/mL)(100) = 114 IU/100mL
  • T = 114 IU/100mL of Vitamin A

Example 2:

In this example, 24mL of a 205,000 IU/mL vitamin D solution is used to produce 500mL of diluted vitamin solution. The vitamin pump speed is 9840 IU/mL and the flow rate of milk is 64L/minute. Determine the theoretical concentration of vitamin D.

First, determine the Concentration of the diluted Vitamin solution (10.3.1.1):

Cd = (Cp × Vp) / Vd

Cd = (205,000 IU/mL × 24mL) / 500mL = 9840 IU/mL

Then use this value for "C" in the formula T= CP/ F

  • T = CP/F
  • T = (9840 IU/mL × 2.6mL/minute) / 64,000mL/minute
  • T = (25584 IU/minute) / 64,000mL/minute
  • T = 0.400 IU/mL of Vitamin D

This needs to be converted to IU/100mL. Do this by multiplying your calculated T value by 100.

Therefore:

  • (0.400 IU/mL)(100) = 40.0 IU/100mL
  • T = 40 IU/100mL of Vitamin D

10.3.2 Calculated Values

Information required:

  • CV = Calculated level of vitamins in final product (IU/100mL)
  • C = Vitamin Premix Concentration (or concentration of diluted Vitamin solution) (IU/mL)
  • U = Total volume of Vitamin Premix used (or volume of diluted Vitamin solution) (mL)
  • Q = Total Quantity of fortified finished product (mL)Footnote 9

Equation used:

CV = CxU/Q

Example 1:

  • CV = Calculated level of vitamins in final product (IU/100mL)
  • C = 50,000 IU/mL (of Vitamin A)
  • U = 64.8mL
  • Q = 1810L (1810L × 1000mL/L = 1,810,000mL)
  • CV = CxU/Q
  • CV = (50,000 IU/mL × 64.8mL) / 1,810,000mL
  • CV = 3240000 IU / 1,810,000mL
  • CV = 1.79 IU/mL of Vitamin A

This needs to be converted to IU/100mL. Do this by multiplying your calculated T value by 100.

Therefore:

  • (1.79 IU/mL)(100) = 179 IU/100mL
  • T = 179 IU/100mL of Vitamin A

Example 2:

In this example, 126.2mL of a 205,000 IU/mL vitamin D solution is used to produce 7.6L of diluted vitamin D solution.

First, determine the Concentration of the diluted Vitamin solution (10.3.1.1):

  • Cd = (Cp × Vp) / Vd
    • = (205,000 IU/mL × 126.2mL) / 7600mL
    • = (258710000 IU) / 7600mL
    • = 3404.1 IU/mL

Then use this value for "C" in the formula CV = CU/Q.

  • CV = Calculated level of vitamins in final product (IU/100mL)
  • C = 3404.1 IU/mL
  • U = 7560mL
  • Q = 60,586L (60,586L × 1000mL/L = 60,586,000mL)
  • CV = CU/Q
  • CV = (3404.1 IU/mL × 7560mL) / 60,586,000mL
  • CV = 25734996 IU / 60,586,000mL
  • CV = 0.425 IU/mL of Vitamin D

This needs to be converted to IU/100mL. Do this by multiplying your calculated T value by 100.

Therefore:

  • (0.427 IU/mL)(100) = 42.7 IU/100mL
  • T = 42.7 IU/100mL of Vitamin D

10.3.2.1 Determining the Volume of Vitamin Premix Prepared by Weighing

Sometimes, solutions may be prepared by weighing the premix instead of using a volumetric measure. In order to convert the weight of premix to a volume, you need to know the Specific Gravity (density) of the premix.

Information required:

  • V = Amount of vitamin premix used (mL)
  • M = Weight of premix used (g)
  • SG = Specific Gravity (density) of premix (g/mL)

Formula Used:

  • V = M/SG

Example:

  • V = x (mL)
  • M = Weight of premix used = 3.5g
  • SG = Specific Gravity (density) of premix = 1.045g/mL

Use the formula V = M/SG = 3.5g/(1.045g/mL) = 3.35mL of vitamin premix used

10.4 Degree-hours Calculation

Degree-Hours are the product of time (h) at a particular temperature multiplied by the temperature (°C) in excess of the critical temperature at which the growth of a particular microbiological pathogen effectively begins. Degree-Hours are calculated for each temperature used in the process.

Certain strains of the bacteria Staphylococcus aureus are capable of producing a highly heat stable toxin that causes illness in humans. Above a critical temperature of 15.6°C, Staphylococcus aureus multiplication and toxin production can take place. This multiplication and resultant toxin production ceases once a pH of 5.3 is reached. Processors are consequently required to control this hazard by verifying that their product attains a pH of 5.3 within pre-defined degree-hours limits. The examples presented in this section will reflect the control of Staphylococcus aureus; however, the methodology is the same for other microbiological pathogens. The appropriate Guideline table for the microbiological pathogen in question must be used.

10.4.1 Fermentation at a constant temperature (Constant Temperature Process)

When fermentation is done at a constant temperature, operators can either use the Guidelines table or the calculation method to determine the degree-hours limits and maximum time for fermentation at a given room temperature.

10.4.1.1 Determining the Maximum Degree/Hours Using the Guideline Table

Use the following table (Table 7) to determine the maximum degree-hours for fermentation room temperatures to control Staphylococcus aureus:

  1. less than 33°C
  2. between 33°C and 37°C
  3. in excess of 37°C
Table 7 – Degree-hours guidelines for the control of Staphylococcus aureus.
Degrees-hours limit for the corresponding temperatureFermentation Room Temperature (°C)Maximum Allowed Hours to Achieve a pH of 5.3 (Based on Guideline)
665 20 150.0
665 22 103.4
665 24 78.9
665 26 63.8
665 28 53.6
665 30 46.2
665 32 40.5
555 33 31.8
555 34 30.1
555 35 28.6
555 36 27.2
555 37 25.9
500 38 22.3
500 40 20.5
500 42 18.9
500 44 17.6
500 46 16.4
500 48 15.4
500 50 14.5

Answers:

  1. 665 degree-hours when the highest fermentation temperature is less than 33°C.
  2. 555 degree-hours when the highest fermentation temperature is between 33°C and 37°C.
  3. 500 degree-hours when the highest fermentation temperature is greater than 37°C.

Note: The degree-hours above depend upon the highest temperature in the
fermentation process prior to the time that a pH of 5.3 or less is attained.

10.4.1.2 Determining the Maximum Degree-Hours Using the Calculation Method

Information required:
  • Degrees in excess of the critical temperature for microbial pathogen growth (e.g. 15.6°C for Staphylococcus aureus)
  • Hours required to reach pH necessary to stop microbial pathogen growth (e.g. 5.3 for Staphylococcus aureus)
Steps:
  1. Subtract the critical temperature required for microbial pathogen growth from the temperature of the fermentation room
  2. Determine the number of hours required at the above temperature necessary to reach the critical pH value
  3. Multiply steps 1 and 2
Examples:

All examples will be for the control of Staphylococcus aureus. You will determine whether each process meets the guideline presented in the preceding table.

Process A

The fermentation room temperature is a constant 26°C. It takes 55 hours for the pH to reach 5.3.

  • Degrees above 15.6°C: 26°C - 15.6°C = 10.4°C
  • Hours to reach pH of 5.3: 55 h
  • Degree-Hours calculation: (10.4°C) × (55 h) = 572 degree-hours

The corresponding degree-hours limit (less than 33°C) is 665 degree-hours.

Conclusion: Process A meets the guideline because its degree-hours is less than the limit.

Process B

The fermentation room temperature is a constant 35°C. It takes 40 hours for the pH to reach 5.3.

  • Degrees above 15.6°C: 35°C - 15.6°C = 19.4°C
  • Hours to reach pH of 5.3: 40 h
  • Degree-Hours calculation: (19.4°C) × (40 h) = 776 degree-hours

The corresponding degree-hours limit (between 33°C and 37°C) is 555 degree-hours.

Conclusion: Process B does not meet the guideline because its degree-hours exceed the limit.

10.4.2 Fermentation Done at Different Temperatures (Variable Temperature Processes)

When the fermentation takes place at various temperatures, each step in the progression is analysed for the number of degree-hours it contributes using the calculations outlined in the preceding section. The degree-hours are then totalled for all the steps, and compared to the guideline recommendation (Table 8) for the highest temperature reached during fermentation.

Examples:

All examples will be for the control of Staphylococcus aureus.

Process C

It takes 35 hours for product to reach a pH of 5.3 or less. Fermentation room temperature is 24°C for the first 10 hours, 30°C for second 10 hours and 35°C for the final 15 hours.

Table 8: Times and Temperatures –Process C
Hours Fermentation Room temperature (°C) Critical Temperature Adjustment (°C) Degrees above 15.6°C Degree-hours
10 24° (24° - 15.6°) = 8.4° 84
10 30° (30° - 15.6°) = 14.4° 144
15 35° (35° - 15.6°) = 19.4° 291
Total: 519

The highest temperature reached = 35°C

The corresponding degree/hour limit = 555 degree-hours (between 33°C and 37°C)

Conclusion: Process C meets the guideline because its degree-hours is less than the limit.

Process D

It takes 38 hours for product to reach a pH of 5.3 or less. Fermentation room temperature is 24°C for the first 10 hours, 30°C for second 10 hours and 37°C for the final 18 hours.

Table 9: Times and Temperatures – Process D
Hours Fermentation room temperature (°C) Critical Temperature Adjustment (°C) Degrees above 15.6°C Degree-hours
10 24° (24° - 5.6°) = 8.4° 84
10 30° (30° - 15.6°) = 14.4° 144
18 37° (37° - 15.6°) = 21.4° 385.2
Total: 613.2

The highest temperature reached = 37°C

The corresponding degree-hour limit = 555 (between 33°C and 37°C)

Conclusion: Process D does not meet the guidelines because its degree-hours exceed the limit.

10.5 Common Food Additives

10.5.1 Nitrite and Nitrate Additives

According to the Food and Drug Regulations, the combination of sodium/potassium nitrite and the combination of sodium/potassium nitrate should not exceed 200 ppm each (200 ppm + 200 ppm)

Note: should not exceed 120 ppm for side bacon

10.5.1.1 Determining How Much Nitrite/Nitrate to Add to Achieve a Specific Concentration

Information required:

  • Quantity of product to which the nitrite or nitrate will be added
  • Maximum quantity of nitrite or nitrate for the above quantity of product

Useful conversions:

  • 1 kg = 1000 g
  • 200 ppm = 0.02%
  • 1 g of product may contain a maximum 200 μg sodium nitrate/nitrite
  • 1 kg of product may contain a maximum 200 mg (0.2 g) of sodium nitrate/nitrite

Formula:

Desired ppm combined nitrates/nitrates (g/kg) = [Amount of combined nitrates/nitrates (g)] / Batch size (kg)

Example:

A company wants to add 200 ppm sodium nitrite to 250 kg of sausage mix. How much sodium nitrite should they add?

What we know:

Desired ppm = 200 ppm = 200 mg/kg = 0.200 g/kg

Final batch size = 250 kg sausage mix

  • Desired ppm sodium nitrate (g/kg) = Amount of sodium nitrate (g) / Batch size (kg)
  • [0.200 g sodium nitrate / 1 kg Final mixture] = [x g sodium nitrate / 250 kg Final mixture]
  • (0.200)(250) = × g sodium nitrate
  • 50 = x g sodium nitrate

Therefore, the company should add 50 g of sodium nitrate to 250 kg of sausage mix to achieve 200 ppm sodium nitrate in the sausage mix.

10.5.1.2 Determining How Much Product to Add to Achieve a Specific Concentration

For this style of question, you may use the equation:

  • C1M1 = C2M2
  • C1 = initial concentration
  • C2 = final concentration
  • M1 = initial mass (or volume)
  • M2 = final mass (or volume)

Example:

According to the company laboratory report, the latest batch of sausage mix weighs 200 kg and contains 340 ppm of nitrate. The manager has decided to add some more untreated sausage mix to the batch to satisfy the Food and Drug Regulations. How much sausage mix must be added to reduce the nitrate concentration to 200 ppm?

  • C1 = initial concentration = 340 ppm nitrate
  • C2 = final concentration = 200 ppm nitrate
  • M1 = initial mass (or volume) = 200 kg sausage mix
  • M2 = final mass (or volume) = 200 kg sausage mix + X (which is additional amount of sausage mix in kg)

Therefore,

  • C1M1 = C2M2
  • (340 ppm nitrate)(200 kg sausage mix) = (200 ppm nitrate)(200 kg sausage mix + x)
  • Divide both sides of the equation by 200 ppm nitrate to reduce the equation to: 340 kilograms sausage mix = 200 kilograms sausage mix + x
  • 340 kg sausage mix = 200 kg sausage mix + x
  • Subtract 200 kilograms sausage mix from both sides of the equation to solve for x
  • 140 kg sausage mix = x

10.5.2 Sulphur Dioxide (SO2)

According to the Food and Drug Regulations, the permitted concentration of sulphur dioxide (SO2) in certain products is 200 ppm. Since SO2 is a gas, and consequently difficult to handle, industry uses various salts of sulphurous acid:

Table 10: Sulphur Dioxide (SO2)
Molar Weight SO2 Content Multiplier
Sulphur dioxide (SO2) 64.07 100% 1.0
Sodium sulfite (Na2SO3) 126.05 50.8% 1.97
Sodium bisulfite (NaHSO3) 104.07 61.5% 1.63
Sodium metabisulfite (Na2S2O5) 190.12 67.4% 1.48

In order to determine how much of any given salt is required for a final SO2 concentration, you must:

  1. determine the amount of SO2 needed for that concentration and then
  2. multiply your result by the appropriate multiplier (see preceding Table) for the salt you wish to use.

Example:

A company wishes to achieve a final SO2 concentration of 150 ppm in 250 kg of product. How much NaHSO3 must they add?

150 ppm SO2 = 150 mg SO2/kg product

Therefore,

  1. 150 mg SO2 / 1 kg product = x mg SO2 / 250 kg product

    (150 mg SO2)(250 kg product) = (x mg SO2)(1 kg product)

    [(150 mg SO2)(250 kg product)] / (1 kg product) = [(x mg SO2)(1 kg product)] / (1 kg product)

    x = 37500 mg

    x = 37.5 g SO2

  2. By applying the multiplication factor from the table above:

    37.5 g SO2 × 1.63 = 61.1 g NaHSO3

Therefore, 61.1 g NaHSO3 must be added in order to achieve 150 ppm SO2.

10.6 Allergens

When an undeclared allergen is found in a food, the severity of consequences is determined by calculating the concentration of allergenic protein in the food (usually in parts per million (ppm) (mg/kg)) and amount of allergenic protein in a serving size of that product (mg protein/serving size). The results of this calculation will be used during the course of a Health Risk Assessment (HRA) to help determine the extent to which human health may be affected. This information will further guide the course of action taken by the CFIA. For more information on this topic, consult Chapter 7 of the CFIA's Food Allergy Reference Manual.

10.6.1 Example calculations for protein concentration in finished goods

Example:

During the course of an inspection, the inspector discovers that there are undeclared allergens in 15g packages of dry soup mix. The manufacturer has failed to indicate in the ingredient listing that there is skim milk powder in the product.

The inspector learns that skim milk powder is present as 5% of the dry soup mix. The specification sheet from the manufacturer of the skim milk powder shows that the skim milk powder is 11% protein.

What is the allergen concentration per serving of this product?

What we know:

  • package size (one serving) = 15 g
  • amount of skim milk powder in soup mix = 5% or 5 g skim milk powder/100 g soup mix
  • amount of milk protein in skim milk powder = 11% or 11 g milk protein/100 g skim milk powder
  • 1 ppm = 1 mg allergenic protein/1 kg product
  • 1% = 10,000 ppm

Method 1:

To determine the amount of skim milk powder in one serving, multiply the serving size by the percentage skim milk powder in the soup mix

  • 15 g soup mix × (5 g skim milk powder / 100 g soup mix) = 0.75 g skim milk powder

To determine the amount of milk protein (from the skim milk powder) in one serving, multiply the amount of skim milk powder in one serving by percentage milk protein in the skim milk powder. This is the amount of milk protein that a consumer will be exposed to from serving of the soup

  • 0.75 g skim milk powder × (11 g milk protein / 100 g skim milk powder) = 0.0825 g milk protein

To determine the concentration (ppm) of the milk protein in the dry soup mix, convert the amount of milk protein to milligrams (mg) and the amount of the serving size to kilograms (kg). Divide the amount of milk protein (in mg) in one serving by the serving size (in kg). This equals the protein concentration in mg allergenic protein per kg product, or ppm.

  • 82.5 mg milk protein / 0.015 kg soup mix = 5500 mg milk protein / 1 kg soup mix

Therefore, the concentration of the milk protein in the dry soup mix is 5500 ppm.

Method 2:

To determine the percent of the dry soup mix that is undeclared milk protein, multiply the percent amount of skim milk powder in the product by the percent milk protein in the skim milk powder: 5% x 11% = 0.05 x 0.11 = 0.0055 = 0.55%

Since we know that 1% = 10,000 ppm:

  • 1% / 10,000 ppm = 0.55% / X ppm
  • X = 5500 ppm

To determine the amount of milk protein in one serving, multiply the concentration in ppm (which is mg allergenic protein/kg product) by the serving size in kg:

  • 5500 ppm x 0.015 kg = 82.5 mg milk protein in one serving of soup mix
Date modified: