Name_____________________
Period________
Show all setups and calculations on a separate piece of paper.
1) What is the empirical formula of a compound containing 60.0% sulfur and 40.0%
oxygen by mass?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
60% S =
60gS
1molS
32.066gS
= 1.8711 molS
40% O =
40gO
1molO
15.9994gO
= 2.500 molO
(Divide by smallest)
(Multiply till whole)
S1.8711 O2.500
x3 =
1.8711
=
S1O1.333
1.8711
S3O4
2) A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen.
What is the empirical formula of this compound?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
23.3% Mg =
23.3gMg
1molMg
24.3050gMg
30.7% S =
30.7gS
1molS
32.066gS
46.0% O =
46.0gO
1molO
15.9994gO
(Divide by smallest)
Mg0.9587 S0.9574 O2.875=
0.9574
.09574
.09574
= 0.9587 molMg
= 0.957 molS
= 2.875 molO
(Multiply till whole)
Mg1S1O3
x1 =
MgSO3
3) What is the empirical formula for a compound containing 38.8% carbon, 16.2%
hydrogen and 45.1% nitrogen?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
38.8% C =
38.8gC
1molC
12.0107gC
= 3.23 molC
16.2% H =
16.2gH
1molH
1.00794gH
= 16.07 molH
45.1% N =
45.1gN
1molN
14.00674gN
= 3.22 molN
(Divide by smallest)
(Multiply till whole)
C3.23 H16.07N3.22=
x1 =
3.22
3.22
C1H5N1
3.22
CH5N
4) A sample of an oxide of nitrogen is found to contain 30.4% nitrogen. What is its
empirical formula?
Oxides are oxygen containing compounds. This one is made of Nitrogen and Oxygen:
30.4% N =
30.4gN
1molN
14.00674gN
= 2.17 molN
69.6% O =
69.6gO
1molO
15.9994gO
= 4.35 molO
(Divide by smallest)
N2.17 O4.35
2.17
2.17
=
N1O2
(Multiply till whole)
x1 =
NO2
5) A sample of an oxide of arsenic is found to contain 75.74% arsenic. What is its
empirical formula?
Oxides are oxygen containing compounds. This one is made of Arsenic and Oxygen:
75.74% As =
75.74gAs
1molAs
74.92160gAs
= 1.01 molAs
24.26 O =
24.26gO
1molO
15.9994gO
= 1.52 molO
(Divide by smallest)
As1.01 O1.52
1.01
=
(Multiply till whole)
As1O1.51 x 2 =
1.01
As2O3
6) What is the empirical formula for a compound containing 26.57% potassium, 35.36%
chromium, and 38.07% oxygen?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
26.57% K =
26.57gK
1molK
39.0983gK
= 0.6796 molK
35.36% Cr =
35.36gCr
1molCr
51.9961gCr
= 0.6801molCr
38.07% O =
38.07gO
1molO
15.9994gO
= 2.38 molO
(Divide by smallest)
K0.6796 Cr0.6801O2.38=
0.6796
0.6796
0.6796
(Multiply till whole)
K1Cr1O3.5
x2 =
K2Cr2O7
7) What is the empirical formula of a compound comprised of 1.8% hydrogen, 56.1%
sulfur and 42.1% oxygen?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
1.8% H =
1.8gH
1molH
1.00794gH
= 1.786 molH
56.1% S =
56.1gS
1molS
32.066gS
= 1.750 molS
42.1% O =
42.1gO
1molO
15.9994gO
= 2.63 molO
(Divide by smallest)
H1.786 S1.750 O2.63=
1.750
1.750
(Multiply till whole)
H1.02S1O1.50
x2 =
1.750
H2S2O3
8) A borane is a compound containing only boron and hydrogen. If a borane is found to
contain 88.45% boron, what is its empirical formula?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
88.45% B =
88.45gB
1molB
10.811gB
= 8.181 molB
11.55 H =
11.55gH
1molH
1.00794gH
= 11.459 molH
(Divide by smallest)
B8.181 H11.459
8.181
11.459
=
(Multiply till whole)
B1H1.40 x 5 =
B5H7
9) Find the empirical formula for a compound containing 40.6% carbon, 5.1% hydrogen,
and 54.2% oxygen.
% to mass  Mass to moles  divide by smallest  Multiply till whole!
40.6% C =
40.6gC
1molC
12.0107gC
= 3.380 molC
5.1% H =
5.1gH
1molH
1.00794gH
= 5.060 molH
54.2% O =
54.2gO
1molO
15.9994gO
= 3.388 molO
(Divide by smallest)
C3.380 H5.060O3.388=
3.380
3.380
(Multiply till whole)
C1H1.5N1
x2 =
3.380
C2H3O2
10) What is the empirical formula of a compound containing 47.37% carbon, 10.59%
hydrogen and 42.04% oxygen?
% to mass  Mass to moles  divide by smallest  Multiply till whole!
47.37% C =
47.37gC
1molC
12.0107gC
= 3.944 molC
10.59% H =
10.59gH
1molH
1.00794gH
= 10.507 molH
42.04% O =
42.04gO
1molO
15.9994gO
= 2.628 molO
(Divide by smallest)
C3.944 H10.507O2.628=
2.628
2.628
2.628
(Multiply till whole)
C1.5H3.998O1
x2 =
C3H8O2