Stoichiometry Problem Solving Study Guide
Balance an equation
1. List all the types of atoms under the yields arrow
2. Count the amount of atoms on each side
3. Add coefficients in front of the compounds to match the amount of atoms
4. Repeat until all the atoms match on each side
Find the molar mass of a compound (have to use a periodic table)
add up all the atomic weights of the compound (the number on the periodic table)
H2 O
H =1 x 2 = 2
O = 16 x 1 = 16
16 + 2 = 18
Water “weighs” 18 g/ mol
Convert Particles (anything, like atoms, molecules) to moles
Divide by Avogadro's Number (6.02 x 10 23)
Convert Moles to particles
Multiply by Avo's number
Convert Mass to moles
Divide by the molar mass
Example: How many moles are in 15.6g of Na? 15.6 / 22.9 =0.681mol Na
Convert moles to grams
Multiply by the molar mass
Example: 0.005 mol of H2O is how many grams? 0.005 mol H2Ox 18g/mol = 0.09 g H2O
Convert liters of a gas (at STP) to moles
1. Convert to L if not already in liters (if mL divide by 1000 first)
2. Divide by 22.4 example: How many moles are in 215mL of an unknown gas? 215÷ 1000 ÷ 22.4 =0.00959
Convert moles to liters of a gas
Multiply by 22.4
example: How many moles in 0.33mol of a He(g)? 0.33 x 22.4 = 7.4L
Convert liters of a liquid to moles
Multiply by the molarity
example: How many moles are in 3L of a 2M solution? 3X2= 6 moles
Calculate Percent Composition
1.Find the total molar mass
2. Find the mass of each element
3. Divide mass of element/ total mass
4. Convert to percentage (multiply by 100)
Example: What is the % composition of H2 O?
H2 O = 18g/ mol (total)
H=1x2=2
2 ÷ 18 = 11.1%
O = 16 x1 =16
16 ÷ 18 = 88.9%
Water is 11.1% H and 88.9% O
Finding the Empirical formula from the Percent Composition
1. Convert both elements to moles (÷ by molar mass)- if a percent, just change % to g
2. find the smallest whole number ratio
a. divide by the smallest number
b. if you don't get whole numbers (or close) multiply by a whole number until you get all whole
Example What is the formula of a substance that is 15.8% H and 84.2% O?
15.8 H ÷ 1 = 15.8 mol H ÷ 5.625 =3
84.2 O ÷ 16 = 5.625 mol O ÷5.625 =1
Answer: H3O
Finding the Molecular Formula from the Empirical Formula and Molar Mass
1. Divide the total molar mass by the molar mass of the empirical formula
2. Multiply the empirical formula’s subscripts by that number
Example: What is the molecular formula of a compound with the empirical formula of CH 3O and
has a molar mass of 93g/mol?
Empirical Formula
CH3O = 31g/mol
Total Molar Mass
93g/mol
93 ÷ 31 = 3
Answer= C3H9O3
number
Finding the Formula of a hydrate
1. convert grams of anhydrous compound to moles (divide by molar mass)
2. find grams of water by subtracting the original mass from the mass you get afterwards
3. convert gram s water to moles (divide by 18-the molar mass of water)
4. divide each by the moles of the anhydrous compound
You heat 5.56g of a hydrate (MgSO4 XH2O) and the anhydrous product has a mass of 2.75, what
is the formula of the hydrate?
MgSO4 = 120.3g/mol
2.75 ÷ 120.3= 0.0223mol ÷0.0223 = 1
2.81g ÷ 18 = 0156mol ÷0.0223= 7
Answer = MgSO4 7H2O
Mass to Mass conversions
1. balance the equation
2. convert grams to moles of what is known (divide by molar mass)
3. convert moles to moles
a. multiply by the coefficient of what you are solving for
b. divide by the coefficient of the original compound
4. convert moles to grams by multiplying by the molar mass
Example: If you react hydrogen gas and oxygen gas it will produce water. If you started with 25.2g of H 2
gas, how many grams of water would you make?
balanced equation: 2H2 + O2  2H2O
grams to moles 25.2g H2 ÷ 2g.mol = 12.6mol
moles to moles H2 12.6 x 2 ÷ 2 = 12.6mol H2O
moles to grams 12.6 x 18= 226.8gH2O (the answer is 226.8g)
Finding the limiting reactant/ excess reactant
1. balance the equation
2. convert both substances to moles
3. divide each by its own coefficient
4. the larger number is what is in excess, the smaller number is what is the limiting reactant
Finding the amount of excess reactant (what is leftover)
(must recalculate based on what is the limiting reactant-determined by steps 1-4)
5. erase/ignore the data of the excess
6. convert moles of limiting reactant to moles of excess reactant
a. multiply by the coefficient of the excess reactant
b. divide by the coefficient of the limiting reactant
7. convert moles of limiting reactant to grams (or another useful measurement)
8. subtract (the actual amount of excess product – the amount that reacted(determined by step 7))
Example: If you react hydrogen gas and oxygen gas it will produce water. If you started with 50.0g of H 2
and 50.0g of O2 gas, which is the limiting reactant? How much excess do you have?
balanced equation: 2H2 + O2  2H2O
50.0g H2 ÷ 2 = 25mol H2 ÷2 = 12.5
50.0g O2 ÷ 32 = 1.56mol O2 = 0.78
H2 is in excess, O2 is limiting
1.56 mol O2 x 2 ÷ 1= 3.12 mol H2
3.12 x 2g/mol = 6.24g H2
50.0-6.24 = 42.8 g H2 in excess
Finding Percent Yield
1. Find the error ( error = |experimental data – theoretical data| )
2. find percent error (% error = error/ theoretical data x100)
3. subtract % error from 100
Example: In a chemical reaction, I produced 23g of NaCl. I should have made 30g. What is the
yield?
|23-30| = 7
7/30 = 0.23
0.23 x 100 = 23 %
100-23 = 77% yield