Calculate empirical formula when given mass data

Return to Mole Table of Contents

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data

Determine the formula of a hydrate


Example #1: A sample of copper metal weighing 2.50 g is heated to form an oxide of copper. The final mass of the oxide is 3.13 g. Determine the empirical formula of the oxide.

Solution:

1) Determine mass:

Cu ---> 2.50 g
O ---> 3.13 g − 2.50 g = 0.63 g

2) Determine moles:

Cu ---> 2.50 g / 63.546 g/mol = 0.03934 mol
O ---> 0.63 g / 16.00 g/mol = 0.039375 mol

This is a 1:1 molar ratio between Cu and O.

3) Write the empirical formula:

CuO

Example #2: On analysis, a compound with molar mass 60 g/mol was found to contain 12.0 g of carbon, 2.0 g of hydrogen and 16.0 g of oxygen. What is the molecular formula of the compound? (no calculator allowed!)

Solution:

12.0 g carbon is about 1 mole of carbon; 2.0 g of H is about 2 moles and 16.0 g O is about one mole. So the empirical formula is CH2O

The molecular weight of this molecule is 30 g/mole, so you divide 60/30 to find how many times you must multiple your empirical formula. The answer is 2 times the above empirical formula, so the molecular formula is C2H4O2


Example #3: A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The following data was collected:

Mass of crucible: 38.26 g
Mass of crucible and iridium: 39.52 g
Mass of crucible and iridium oxide: 39.73 g

Solution:

1) Get mass of each element:

Ir ---> 39.52 − 38.26 = 1.26 g
O ---> 39.73 − 39.52 = 0.21 g

2) Get moles of each element:

Ir ---> 1.26 g / 192.217 g/mol = 0.0065551 mol
O ---> 0.21 g / 16.00 g/mol = 0.013125 mol

3) Look for smallest whole-number ratio:

Ir ---> 0.0065551 / 0.0065551 = 1
O ---> 0.013125 / 0.0065551 = 2

Empirical formula = IrO2

By the way, this problem ignores any errors that might be produced by the Ir reacting with the nitrogen in the air.


Example #4: A compound contains 16.7 g of iridium and 10.3 g of selenium, what is its empirical formula?

Solution:

1) Moles:

Ir ---> 16.7 / 192.217 = 0.086881
Se ---> 10.3 / 78.96 = 0.130446

2) Seek lowest whole-number ratio:

Ir ---> 0.086881 / 0.086881 = 1
Se ---> 0.130446 / 0.086881 = 1.5

3) Multiply 1 : 1.5 ratio by two:

2 : 3

Ir2Se3


Example #5: A 1.000 g sample of red phosphorus powder was burned in air and reacted with oxygen gas to give 2.291 g of a phosphorus oxide. Calculate the empirical formula and molecular formula of the phosphorus oxide given the molar mass is approximately 284 g/mol.

Solution:

1) Calculate moles of P and O:

P ---> 1.000 g / 30.97 g/mol = 0.032289 mol
O ---> 1.291 g / 16.00 g/mol = 0.0806875 mol

2) Determine lowest whole-number ratio:

P ---> 0.032289 mol / 0.032289 mol = 1
O ---> 0.0806875 mol / 0.032289 mol = 2.50

3) Determine empirical formula:

P2O5

3) Determine molecular formula:

The "empirical formula weight" = 141.943

284 / 142 = 2

P4O10


Example #6: A sample of magnetite contained 50.4 g of iron and 19.2 g of oxygen. Calculate the empirical formula.

Solution:

1) Convert grams to moles:

Fe ---> 50.4 g / 55.845 g/mol = 0.9025 mol
O ---> 19.2 g / 16.0 g/mol = 1.2 mol

2) Seek smallest whole-number ratio:

Fe ---> 0.9025 / 0.9025 = 1
O ---> 1.2 / 0.9025 = 1.33 <--- a common error is to round this off to 1

3) A ratio that involves a third (like the 1.33 just above or something like 2.67) can be thought of as fractions with a denominator of 3. Like this:

Fe ---> 3/3
O ---> 4/3

4) Multiply by three to get to whole numbers:

Fe ---> 3
O ---> 4

The empirical formula is Fe3O4


Example #7: If one molecule with the empirical formula C3H7 has a mass of 1.428 x 10¯22 g, determine the molecular formula of the compound.

Solution:

1) Determine mass of one mole:

(1.428 x 10¯22 g/molecule) (6.022 x 1023 molecules/mol) = 86.0 g/mol

2) Determine "empirical formula weight:"

C3H7 = 43.1

3) Determine molecular formula:

86.0 / 43.1 = 2

C6H14


Example #8: What is the empirical formula and molecular formula for lactic acid if the percent composition is 40.00% C, 6.71% H, 53.29% O, and the approximate molar mass is 90 g/mol?

Solution:

1) Assume 100 g of the compound is present. This turns percents into mass.

2) Calculate moles:

C ---> 40.00 g / 12.011 g/mol = 3.33
H ---> 6.71 g / 1.008 g/mol = 6.66
O ---> 53.29 g / 16.00 g/mol = 3.33

3) Divide by lowest value:

C ---> 3.33 / 3.33 = 1
H ---> 6.66 / 3.33 = 2
O ---> 3.33 / 3.33 = 1

Empirical formula = CH2O

4) Empirical formula weight

12 + 2 + 16 = 30

5) What is the molecular formula?

90/30 = 3 (there are three "units" of the empirical formula in the molecular formula)

C3H6O3


Example #9: A 0.338 g sample of antimony was completely reacted with 0.295 g of chlorine gas to form antimony chloride. Determine the empirical formula of the antimony chloride.

Solution:

Sb ---> 0.338 g / 121.760 g/mol = 0.002776 mol
Cl ---> 0.295 g / 35.453 g/mol = 0.008321 mol

0.008321 mol / 0.002776 mol = 3

SbCl3

Note the use of the atomic weight of chlorine, not the molecular weight of Cl2. We are interested in determining how many Cl atoms bond per one Sb atom, not how many Cl2 molecules bond with one Sb. If you use 70.906 g/mol, you get 1.5 chlorine to one Sb (and forgetting that it's 1.5 Cl2 molecules) you conclude the formula is Sb2Cl3.


Example #10: When the element antimony, Sb, is heated with excess sulfur, a reaction occurs to give a compound containing only antimony and sulfur. On further heating, excess sulfur is burnt off forming gaseous sulfur dioxide, SO2, and the substance left is pure compound of antimony and sulfur. In one experiment, 2.435 g of antimony was used, and the mass of the pure compound of antimony and sulfur was found to be 3.397 g. What is the empirical formula of the antimony-sulfur compound?

Solution:

1) Calculate mass of sulfur reacted:

3.397 − 2.435 = 0.962 g

2) Calculate moles:

Sb ---> 2.435 g / 121.76 g/mol = 0.0200 mol
S ---> 0.962 g / 32.065 g/mol = 0.0300 mol

3) Divide through by smallest:

0.02 / 0.02 = 1
0.03 / 0.02 = 1.5

4) Multiply by 2 to get whole number ratio

Sb2S3

Comment: you can see the 0.2 to 0.3 ratio is the same as a 2 to 3 ratio and arrive at the empirical formula that way. The "divide by smallest" is the classic way to solve problems of this type.


Example #11: 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.


Example #12: 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

Example #13: 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


Example #14: 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

Example #15: 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.


Example #16: 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

Example #17: 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

Example #18: 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

Example #19: 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


Bonus Example: Which mass ratio would be obtained if methanol is analyzed?

     C    H    O  
A.3g1g4g
B.3g2g8g
C.6g8g16g
D.12g8g32g

Solution:

1) The formula for methanol is:

CH3OH

2) This means the following atom ratio:

one carbon : four hydrogen : one oxygen

3) Use the atomic masses of each element to obtain the mass ratio for the above atom ratio:

12 grams C : 4 grams H : 16 grams O

4) We are looking for an answer that has the weight in the ratio of 12 : 4 : 16. Reduced to lowest terms, that ratio is:

3 : 1 : 4

Answer choice A


Return to Mole Table of Contents

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data

Determine the formula of a hydrate