How to solve empirical formula problem –
If you are given a percentage, you pretend you have 100 g so that each percent becomes a mass.
For example – 62.0% C, 10.4% H, 27.5% O becomes:
62 g C, 10.4 G H, 27.5 g O
Convert all grams to moles by dividing by the molar mass of the element.
Carbon – 62/12.01 = 5.16 moles C /1.175 = 3
Hydrogen - 10.4/1.008 = 10.3 moles H / 1.175 = 6
Oxygen – 27.5/16 = 1.175 moles O / 1.175 = 1
Now divide all numbers by the smallest number – your empirical formula is C3H6O!!!
If you end with something like 3.5 in one of your numbers, then multiply everything by 2.
How to solve stoichiometric problems:
Example 1:
6Li (s) + N2 (g) → 2Li3N (s)
How many moles of N2 are needed to react with 0.500 mol of lithium?
Confirm the equation is balanced –
You already have moles so you can move to determining ratios between the compounds in question.
The ration of Li to N2 is 6:1
So you can set up an equation such as 6/1 = 0.500/x and solve for x!
X = 0.0833
Example 2 – with grams
Automotive air bags inflate when sodium azide decomposes explosively to its constituent elements:
2NaN3 (s) → 2Na (s) + 3N2 (g)
How many grams of sodium azide are required to produce 33.0 g of nitrogen?
1.
2.
3.
Convert grams to moles – 33.0 g/28 g mol-1 = 1.18 mol N2
Use ratios 3 N2: 2NaN3; 3/2 = 1.18/x solve for x. x = 0.79 mol NaN 3
Convert back to grams by multiplying by molar mass of NaN3; 0.79 x 65 = 51.35 g
How to determine % mass in a compound:
Calculate the percentage by mass of nitrogen in Pb(NO3)2
Figure out total molar mass by adding all of the atoms – be sure to understand how the subscripts apply.
For example there is one Pb, 2 N, and 6 O. Add up molar masses – 207.2 + 2(14) + 6(16) = 331.2g/mol
How much N is there? Two N have a mass of 28 g/mol. Divide your mass of element in question by total
mass and multiply by 100.
28/331.2 x 100 = 8.45%
How to solve percent yield problems when given the percent yield:
Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride:
S (s) + 3F2 (g) → SF6 (g)
In a particular experiment, the percent yield is 79.0%. This means that in this experiment, a 7.90 –g
sample of fluorine yields __________ g of SF6. Assume you have excess S.
You treat this like a regular yield problem and first determine theoretical yield. You are limited only by
the amount of fluorine so use fluorine to determine how much SF6 you could theoretically get.
Convert g to moles (divide mass by molar mass) – 7.90g/38g mol-1 = 0.208 mol F2
Use ratios to determine yield of SF6 = 3 F2: 1 SF6
3/1 = 0.208/x (solve for x)
X = .069 mol SF6 (multiply by molar mass of SF6)
0.069 X 146.1 = 10.08 g is THEORETICAL YIELD but you are told you only have 79% yield.
Multiply your actual by .79 (10.08 x .79 = 7.96g)
How to solve percent yield problems – part 2:
Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide:
CaO (s) + H2O (l) → Ca(OH)2 (s)
In a particular experiment, a 5.00-g sample of CaO is reacted with excess water and 6.11 g of Ca(OH) 2 is recovered.
What is the percent yield in this experiment?
1. Convert grams to mol – 5.00 g CaO divided by molar mass: 5/56.1 = .089 mol CaO
2. Determine how much product you could theoretically get by using ratios.
a. 1:1 ratio means you get 0.089 mol product – convert back to grams - .089 x 74.1 = 6.59 g
Ca(OH)2.
b. Percent yield is actual/theoretical x 100
i. 6.11/6.59 = .926 x 100 = 92.6%
How to solve limiting reaction:
Lithium and nitrogen react in a combination reaction to produce lithium nitride:
6Li (s) + N2 (g) → 2Li3N (s)
In a particular experiment, 3.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is
__________ g.
1. Convert to mol – you have 3.5 g of EACH reactant so do it for both.
a. Li – 3.5 g/6.9g mol-1 = 0.51 mol
b. N2 – 3.5 g/28 g mol-1 = 0.125 mol
2. Use ratios to determine the limiting reactant – pick one to start with:
a. Using Li – 6:1 ratio so 6/1 = 0.51/x. solve for x = .085 mol of N2 needed to react this
amount of Li. So you check and see how much N2 you have –(0.125 N2) so you have
excess N2.
b. Using N2 – 1:6 ratio so 1/6 = 0.125/x. solve for x = .75 mol of Li needed to react this
amount of N2. Compare to what you have – you only have 0.51 mol Li so Li must be
limiting reactant.
3. Use the limiting reactant to determine how much product you will get with ratios.
a. 6Li:2Li3N so 6/2=0.51/x. Solve for x. x = .17 mol
b. Convert to grams by multiplying by molar mass = 0.17 x 34.82 (molar mass of Li3N) = 5.9
g is the yield.
How to solve an excess problem with limiting reactants:
Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide:
CaO (s) + H2O (l) → Ca(OH)2 (s)
A 4.50-g sample of CaO is reacted with 4.34 g of H2O. How many grams of water remain after completion of
reaction?
1. Confirm it is balanced.
2. Treat as a regular limiting reaction problem by determining which reactant is limiting.
3. Convert grams to moles of both –
a. 4.50 g CaO would be 4.50/56.1 = 0.0802 mol CaO
b. 4.34 g H2O would be 4.34/18 =0 .241 mol H2O
c. Lucky you – you have a 1:1 ratio!! Yeah!!!! So CaO is obviously the limiting reactant and
you would need 0.0802 mol of H2O to react 0.0802 mol of CaO.
d. That means you have the original amount of water – the amount used left over or 0.241
mol – 0.0802 mol = 0.161 mol water left over.
e. Convert to grams (x 18g/mol) = 2.9 g left or excess
Final Practice problem for good luck!
If 294 grams of FeS2 is allowed to react with 176 grams of O 2 according to the following equation, how many grams
of Fe2O3 are produced?
4 FeS2 + 7 O2 → 2 Fe2O3 + 4 SO 2
Solve this on your own!!!
Check you answer: 195.5 g Fe2O3