Empirical Formulas from % Composition

[Averie Moore 2F]

I was looking at Adila's Week 1 Step Up worksheet and had a question about #2, determining empirical formulas for compounds with percent compositions. When doing this problem, I know I have to look at the periodic table and find the molecular mass to calculate further. For problems like these, am I supposed to assume the sample size is 100g unless a different measure is given?

[Lucas Vu 2G]

Unless a molar mass of the compound is given, assuming the sample to be 100g is the easiest way to set up the calculations to find the empirical formula. By using 100g it's easier to convert the percentages into masses which can then be converted into moles and lastly factored down to the empirical formula of the compound.

[Sophia Henry]

The reason for using "100g" as the go-to sample mass is because it allows us to easily turn a percent composition into knowledge about the makeup of a sample. More specifically, when given a sample that has this % of this substance and this % of this other substance, by pretending that instead of them telling you you have percents of things you can pretend you have a grams amount of those things. Use the % values to make up masses as if the sample is 100g total and for example you can go from 35% C to 35 g carbon and then when you have an amount of some element in grams this changes everything. Thanks to the periodic table and all of the information it stores we can turn any grams amount of some element into the moles amount of that element. The way you do that is by looking at the periodic table's molar masses it lists. The molar masses are the decimal values that lie beneath atom names. The molar masses are given in g/mol units adn they are essentially unit conversion factors that we can use to determine the amount of moles of something in a sample as long as we know a value for how many grams of that substance there is in the sample. For exmaple with carbon, if we have 35 g of C we can use the molar mass of carbon which is approximately 12 g/mol to perform dimensional analysis calculations to convert grams of carbon to moles of carbon. This relates to the problem you refer to because it shows why it is useful to pretend you have 100 g of sample. You could pretend you have 50 grams of sample and then it would be 35% of the sample being carbon so instead of pretending you have 35 g of carbon in a 100g sample you instead pretend you have (50)(0.35) grams of carbon in the 50 grams of total hypothetical sample.