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Stoichiometry and Basic Chemistry Math

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Total cards: 9•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (9)

#1
Front (Question)

Molecular formula

Back (Answer)

Specify the type and number of each atom in a molecule

#2
Front (Question)

Empirical formula

Back (Answer)

Ratio of atoms in a compound

#3
Front (Question)

Give an example of a a molecular formula and it’s empirical formula

Back (Answer)

Glucose: molecular- C6H12O6 empirical- CH2O

#4
Front (Question)

What is the empirical formula for a 150g sample of 60g carbon, 10g hydrogen, and 80g oxygen?

Back (Answer)

10g C= 5 mol 10g H= 10 mol 80g O= 5 mol C5H10O5 –> CH2O

#5
Front (Question)

Given the empirical formula CH2O, what is the molecular formula if the molar mass of the sample is 180g/mol?

Back (Answer)

C= 12 g/mol H= 1 g X 2= 2 g/mol O= 16 g/mol Total mass of CH2O= 30g 180 g/30g= 6 so, C6H12O6

#6
Front (Question)

What is the amount in grams of each individual element of a 150 g compound that contains 40% carbon, 7% hydrogen, and 53% oxygen?

Back (Answer)

Carbon- 150g X 0.4 = 60 g Hydrogen- 150g X 0.07= 10g Oxygen- 150g X 0.53= 80g then convert to moles On MCAT there is no calculator, so using 100g instead of 150g works too and can see the 1:2:1 ratio

#7
Front (Question)

What is the percent composition by mass of oxygen in a molecule with the empirical formula CH2O?

Back (Answer)

Carbon- 1 X 12g= 12 g/mole Hydrogen- 2 X 1g= 2 g/mol Oxygen= 1 X 16g= 16 g/mol Added together = 30 g/mol C- 12/30= 40% H- 2/30= 7% O- 16/30= 50%

#8
Front (Question)

Identifying limiting reagents

Back (Answer)

The reagent that makes fewer

#9
Front (Question)

Percent yield

Back (Answer)

(Amount of product formed)/ (theoretical yield) X 100 Actual yield: given theoretical yield: calculated