Problem #1: A 4.628-g sample of an oxide of iron was found to contain 3.348 g of iron and 1.280 g of oxygen. What is simplest formula for this compound?

Solution:

1) Determine moles of each element:

iron: 3.348 g / 55.845 g/mol = 0.05995 mol = 0.06 mol
oxygen: 1.280 g / 16.00 g/mol = 0.08 mol

2) We want the lowest whole number ratio between 0.06 and 0.08:

0.06 / 0.06 = 1 <--- think of it as 3/3
0.08 / 0.06 = 1.3333 <--- think of it as 4/3

3) Multiply by 3:

3/3 x 3 = 3
4/3 x 3 = 4

Fe₃O₄

Another way to approach this is to see that 0.06 to 0.08 is a 6 to 8 ratio and then reduce it to 3 to 4.

Problem #2: A sample of magnesium weighing 1.00 gram is burned in excess oxygen to produce an oxide with a mass of 1.66 g. What is the empirical formula of the oxide produced?

Solution:

1) Convert mass to moles:

Mg ---> 1.00 g / 24.305 g/mol = 0.041144 mol
O ---> 0.66 g / 16.00 g/mol = 0.04125 mol

2) Divide through by the smallest value:

Mg ---> 0.041144 / 0.041144 = 1
O ---> 0.04125 / 0.041144 = 1

3) Empirical formula:

MgO

Problem #3: In the analysis of 2.500 g of an unknown compound, it is found that 0.758 g are calcium, 0.530 g are nitrogen and 1.212 g are from oxygen. Find the empirical formula and the identity of this compound (assume the empirical and molecular formulas are the same).

Solution:

1) Determine moles of each element:

0.758 g Ca / 40.0784 g Ca/mol = 0.018913 mol Ca
0.530 g N / 14.00672 g N/mol = 0.037839 mol N
1.212 g O / 15.99943 g O/mol = 0.0757527 mol O

2) Divide by the smallest number of moles:

0.018913 mol Ca / 0.018913 mol = 1.000
0.037839 mol N / 0.018913 mol = 2.001
0.0757527 mol O / 0.018913 mol = 4.005

3) The empirical & molecular formula and identity are:

CaN₂O₄

or more conventionally:

Ca(NO₂)₂

calcium nitrite

Problem #4: After extensive heating, 0.194 g of uranium produced 0.233 g of a compound with oxygen. Determine the empirical formula.

Solution:

1) Moles of uranium:

0.194 g / 238.0289 g/mol = 0.00081503 mol

2) Moles of oxygen:

0.233 g − 0.194 g = 0.039 g

0.039 g / 15.9994 g/mol = 0.0024376 mol

3) Divide by the smaller number of moles:

0.00081503 mol / 0.00081503 mol = 1.000
0.0024376 mol / 0.00081503 mol = 2.991

4) The empirical formula is:

UO₃

Problem #5: A compound of approximate molar mass 123 g/mol contains only carbon, hydrogen, bromine, and oxygen. Analysis reveals that the compound contains 8 times as much carbon as hydrogen by mass. What might its formula be?

Solution:

1) First the carbon & hydrogen:

C ---> 8 g
H ---> 1 g

C ---> 8 g / 12 g/mol = 0.67 mol
H ---> 1 g / 1 g/mol = 1 mol

C ---> 0.67 mol x 3 = 2
H ---> 1 mol x 3 = 3

The CH empirical formula is C₂H₃.

2) We have no information on Br and O, so some guesswork will be involved.

Let us assume 1 Br and 1 O. What is the weight of C₂H₃OBr?

The answer is 122.9487 g/mol

The most reasonable answer to this problem is C₂H₃OBr.

Problem #6: The following results were obtained in an experiment to determine the formula of an oxide of mercury. It decomposed into its elements when heated.

Mass of empty test tube = 15.45 g
Mass of test tube + oxide of mercury = 17.61 g
Volume of oxygen collected at RTP = 120. mL

From the results of the experiment, determine the molecular formula of this oxide of mercury

Solution:

1) Mass of oxide:

17.61 − 15.45 = 2.16 g

2) Determine the moles, then mass of O₂ produced from the decomposition:

RTP has the following values: 25.0 °C and 1.00 atm

(1.00 atm) (0.120 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.0049072 mol

Mass of O₂ ---> (0.0049072 mol) (32.00 g/mol) = 0.157 g

3) Mass of Hg in the oxide:

2.16 − 0.157 = 2.003 g

4) Divide each mass by respective atomic mass :

Hg = 2.003/200.6 = 0.01 mol

O = 0.157/16.00 = 0.01 <--- just in case, remember that 0.157 g of O₂ is also 0.157 g of O

the Hg to O molar ratio is 1:1

5) Empirical formula:

HgO

Problem #7: A compound consists of 0.722 grams of magnesium and 0.278 grams of nitrogen. What is the empirical formula?

Solution:

Determine moles of Mg and N:

Mg ---> 0.722 g / 24.3 g/mol = 0.0297 mol
N ---> 0.278 g / 14.0 g/mol = 0.019857 mol

2) The ratio of moles should be changed to the smallest whole-number ratio. Divide each mole amount by the smallest value:

Mg ---> 0.0297 mol / 0.019857 mol = 1.5
N ---> 0.019857 mol / 0.019857 mol = 1

3) The ratio must be composed of whole numbers:

Mg ---> 1.5 x 2 = 3
N ---> 1 x 2 = 2

4) The empirical formula:

Mg₃N₂

Problem #8: An organic compound contains 21.20% carbon, 5.30% hydrogen and the rest is arsenic and oxygen. From 0.500 g of that compound was obtained 0.802 grams of magnesium ammonium arsenate, MgNH₄AsO₄. Calculate (a) the mass percent each of arsenic and oxygen in that organic compound and (b) its empirical formula.

Solution:

1) Determine the mass percent of As in MgNH₄AsO₄:

molec. weight of MgNH₄AsO₄ ---> 181.2616 u
atomic weight of arsenic ---> 74.922 u

mass percent of As ---> 74.922 u / 181.2616 u = 0.4133363 (leave as decimal value)

2) Determine mass of As in the 0.802 g:

(0.802 g) (0.4133363) = 0.3314957 g

3) Determine the mass percent of As in the organic compound:

0.3314957 g / 0.500 g = 0.66299 = 66.30%

4) Determine the mass percent of oxygen:

100 − (21.20 + 5.30 + 66.30) = 7.20%

5) Start the empirical formula calculation by assuming 100. g of the compound is present. Therefore,

carbon ---> 21.20 g
hydrogen ---> 5.30 g
oxygen ---> 7.20 g
arsenic ---> 66.30 g

6) Determine moles:

carbon ---> 21.20 g / 12.011 g/mol = 1.76505 mol
hydrogen ---> 5.30 g / 1.008 g/mol = 5.25794 mol
oxygen ---> 7.20 g / 16.00 g/mol = 0.450 mol
arsenic ---> 66.30 g / 74.922 g/mol = 0.88492 mol

7) Divide through by smallest:

carbon ---> 1.76505 mol / 0.450 mol = 3.92
hydrogen ---> 5.25794 mol / 0.450 mol = 11.68
oxygen ---> 0.450 mol / 0.450 mol = 1
arsenic ---> 0.88492 mol / 0.450 mol = 1.97

8) The empirical formula is:

C₄H₁₂As₂O

Problem #9: A compound of xenon and fluorine is prepared by reacting 0.526 g of xenon with an excess of fluorine gas. The compound formed weighed 0.678 g. Determine the molecular formula.

Solution:

1) Determine mass of fluorine reacted:

0.678 g − 0.526 g = 0.152 g

2) Determine moles of each element present in the compound:

Xe ---> 0.526 g / 131.293 g/mol = 0.0040063 mol
F ---> 0.152 g / 18.998403 g/mol = 0.0080007 mol

Note use of the atomic weight of F, not the molecular weight of F₂

3) Express the molar ratio in terms of small whole numbers:

Xe ---> 0.0040063 mol / 0.0040063 mol = 1
F ---> 0.0080007 mol / 0.0040063 mol = 2

4) The molecular formula is XeF₂

Problem #10: When samples of Al, Sn, Zn and Pb were heated in oxygen and completely converted to an oxide, the masses increased by 89% (Al), 27% (Sn), 24.5% (Zn), and 7.7% (Pb). (a) Calculate the mass of oxygen that is combined with 5.00 g of each metal. (b) For each metal in (a), use the mass of oxygen combined with 5.00 g of the metal to calculate the empirical formula of the oxide.

Solution to (a):

1) Calculate mass of oxygen reacted

Al---> (5.00 g) (0.89) = 4.45 g
Sn ---> (5.00 g) (0.27) = 1.35 g
Zn ---> (5.00 g) (0.245) = 1.225 g
Pb ---> (5.00 g) (0.077) = 0.385 g

2) Determine empirical formula for aluminum and oxygen:

Al ---> 5.00 g / 26.98 g/mol = 0.18532 mol
O ---> 4.45 g / 16.00 g/mol = 0.278125 mol

Al ---> 0.18532 mol / 0.18532 mol = 1
O ---> 0.278125 mol / 0.18532 mol = 1.5

empirical formula is Al₂O₃

3) Determine empirical formula for tin and oxygen:

Sn ---> 5.00 g / 118.71 g/mol = 0.04212 mol
O ---> 1.35 g / 16.00 g/mol = 0.084375 mol

Sn ---> 0.04212 mol / 0.04212 mol = 1
O ---> 0.084375 mol / 0.04212 mol = 2

empirical formula is SnO₂

4) Determine empirical formula for zinc and oxygen:

Zn ---> 5.00 g / 65.38 g/mol = 0.076476 mol
O ---> 1.225 g / 16.00 g/mol = 0.0765625 mol

Zn ---> 0.076476 mol / 0.076476 mol = 1
O ---> 0.0765625 mol / 0.076476 mol = 1

empirical formula is ZnO

5) The determination of the empirical formula for the lead-oxygen compound is left to the reader.