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Add two packaged foods without averaging their labels

Scale each ingredient to the quantity used, then sum matching nutrient amounts to estimate a mixed snack.

To estimate a snack assembled from two packaged foods, calculate each portion separately and add the matching nutrient amounts. Averaging the two printed labels is usually wrong because the quantities used may differ.

The FDA label-reading guide explains scaling nutrient values with servings. This original example extends the same arithmetic to two fictional ingredients without claiming to reproduce a tested recipe.

Make an ingredient contribution table

Imagine ingredient A lists 6 g protein per 30 g serving. You use 15 g, half a serving, so its contribution is 3 g protein. Ingredient B lists 3 g protein per 40 g serving. You use 60 g, or 1.5 servings, so it contributes 4.5 g.

IngredientAmount usedServing multiplierProtein contributed
A15 g0.53 g
B60 g1.54.5 g
Mixture75 gNot applicable7.5 g

The total is 7.5 g protein in 75 g of mixture, assuming the full weighed quantities are retained. Averaging the original protein figures, six and three, would give 4.5 but would not describe a defined quantity of this snack.

Calculate a mixture concentration only after summing

Divide the mixture total by its total weight: 7.5 ÷ 75 × 100 = 10 g protein per 100 g. The ingredient concentrations are 20 g per 100 g for A and 7.5 g per 100 g for B. Their simple average, 13.75 g, would be inappropriate because B contributes four times as much food weight.

If you need a 25 g portion estimate, a uniform mixture would give 7.5 × 25 ÷ 75 = 2.5 g protein. A handful that contains mostly A may differ, so even mixing is an explicit assumption.

Keep the recipe editable

Store each original label and actual amount used in its own row. When an ingredient or portion changes, recalculate that row before updating the total. Do not change the mixture total directly, because that hides the source of the change.

Use the same named nutrient and unit throughout. Missing nutrient information should remain missing; an absent value is not a valid zero contribution.

Sources

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