CH 115 Fall 2014
Worksheet 18
1. Answer the following pertaining to grams, moles, and molecules.
a). How many grams are in 2.67 moles of sodium sulfate? 379.14 g Na2SO4
b). How many molecules are present in a 198.6 g sample of bromine gas? 7.48 x
1023 molecules Br2
c). How many moles of carbon dioxide are formed if 65.3 g of carbon dioxide gas is
produced when methane reacts in the presence of oxygen gas? 1.48 moles CO2
d). How many atoms are present in 8.46 moles of copper (II) chloride? How many
grams? 5.09 x 1024 moles CuCl2; 1137.4 g CuCl2
e). How many grams are in a 7.295 mole sample of ammonium oxide? 379.34 g
(NH4)2O
2. What is the difference between an empirical formula and a molecular formula?
An empirical formula is the formula for a compound that has the lowest
whole-number ratios for each of the elements. A molecular formula is the
actual formula for the compound based on the molar mass. The empirical and
molecular formulas can be the same.
Ex. C6H12O6 (glucose) – molecular formula  CH2O – empirical formula
3. What are the steps in calculating an empirical formula when given percent
composition of elements in a compound?
1. Assume you have 100 grams of the compound.
2. Calculate how many grams of each element is present in the 100 gram
sample – should just be equal to the percentage values given (that’s why we
assume 100 grams and not 500, etc.)
3. Convert all gram values to mole values.
4. Divide each mole value through by the lowest mole value calculated. Round
if necessary and applicable. Decimals you SHOULD NOT ROUND: .25, .33, .5,
.66, .75.
5. The values remaining are the subscripts you will use in your empirical
formula.
4. The molar mass of nicotine is 162.1 g/mol. It contains 74.0% carbon, 8.7%
hydrogen, and 17.3% nitrogen. Determine nicotine’s empirical formula and
molecular formula.
74.0 g C  6.1667 mol C
8.7 g H  8.7 mol H
17.3 g N  1.236 mol N
Divide through by 1.236.  C = 5, H = 7, N = 1 (all of these numbers were close
enough to approximate).
C5H7N (empirical formula)
To find the molecular formula, calculate the molar mass of the empirical
formula first. In this case, 81 g/mol. Compare this to the molar mass given in
the problem (162 g/mol). The molecular molar mass is twice the empirical
molar mass (162/81 = 2), so the subscripts for the molecular formula should
be twice that of the empirical formula  C10H14N2