### Calculate empirical formula when given mass dataProblems #1 - 10

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 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.