Calculate empirical formula when given mass data
Problems #11 - 20

Ten Examples      Problems #1 - 10      Return to Mole Table of Contents

Determine 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 molecular formula using the Ideal Gas Law

Determine the formula of a hydrate   Hydrate lab calculations


Only the example and problem questions, no solutions


Problem #11: An unknown substance is only made up of 0.2673 g of hydrogen and 4.2327 g of oxygen. Based on these data, determine the possible identity of this compound.

Solution #1:

1) Find the mass percents of this substance:

0.2673 + 4.2327 = 4.5000 g

(0.2673 / 4.5000) x 100 = 5.94%
(4.2327 / 4.5000) x 100 = 94.06%

2) The mass percentages align with hydrogen peroxide (H2O2) and not with water.

Solution #2:

0.2673 g / 1.008 g/mol = 0.2652 mol (of hydrogen)
4.2327 g / 16.00 g/mol = 0.2645 mol (of oxygen)

This is very close to a 1:1 molar ratio, so the empirical formula of this compound is HO. Hydrogen peroxide (H2O2) has an empirical formula of HO. Water (H2O is its empirical formula) does not.


Problem #12: When a 0.192 g sample of phosphous is burned, 0.440 g of a white oxide of phosphorus is obtained as the product. Determine the empirical formula of the oxide product.

Solution:

Subtract the mass of the phosphorus sample from the mass of the oxide of phosphorus that was the product of the reaction. This will give you the mass of oxygen that is contained in the oxide product.

Find the number of moles of phosphorous (P) present in the initial mass of phosphorus sample that reacted, using the atomic weight of phosphorus.

Find the number of the moles of oxygen (O) that are contained in the oxide by using the mass of oxygen in the oxide and the atomic weight of oxygen.

Obtain the lowest integer ratio possible for the relative number of moles of phosphorus present to moles of oxygen present. Do not round your numbers too much.


Problem #13: When 1.66 g of tungsten metal is heated in excess chlorine gas, 3.58 g of a tungsten chloride is produced. Find the empirical formula of the compound produced.

Solution:

1) Determine the amount of chlorine reacted:

3.58 − 1.66 = 1.92 g

2) Convert to moles:

W ---> 1.66 / 183.84 = 0.00903 mol
Cl ---> 1.92 / 35.5 = 0.0541 mol

3) Look for lowest whole-number ratio:

W ---> 0.00903 mol / 0.00903 mol = 1
Cl ---> 0.0541 mol / 0.00903 mol = 6

4) Empirical formula:

WCl6

Comment #1: it occured to me that the writer of the problem could have mistakenly used 71.0 (for Cl2) to back-calculate the problem when it was originally set up. In that case, we have this for the chlorine:

1.92 / 71.0 = 0.027 mol

0.027 mol / 0.00903 mol = 3

WCl3

Comment #2: tungsten hexachloride is an important compound in tungsten chemistry.


Problem #14: A 40.9-mg sample of phosphorus reacts with selenium to form 119.1 mg of the selenide. Determine the empirical formula of this phosphorus selenide.

Solution:

1) Determine mass of Se:

119.1 mg − 40.9 mg = 78.2 mg of Se

2) Get moles of each:

P ---> 0.0409 g / 30.97 g/mol = 0.00132
Se ---> 0.0782 g / 78.96 g/mol = 0.00099

3) Get lowest whole-number ratio:

P ---> 0.00132 / 0.00099 = 1.333 <--- think of it as 4/3
Se ---> 0.00099 / 0.00099 = 1 <--- think of it as 3/3

Multiply by 3 to get a 4 to 3 ratio

4) Empirical formula:

P4Se3

I found a mention of P4Se3 in this article from 1959.


Problem #15: 0.0833 mol of carbohydrate of empirical formula CH2O contain 1.00 g of hydrogen. What is the molecular formula of the carbohydrate?

Solution:

1.00 g of hydrogen is 0.500 mol

0.0833 mol is to 1.00 mol as 0.500 mol is to x

x = 6.00 mol of hydrogen (in 1.00 mol of the carbohydrate)

The empirical formula of CH2O must be scaled up to contain 6.00 moles of H in one mole of the compound.

CH2O times 3 = C3H6O3


Problem #16: What is the empirical formula when 46.4 mg of carbon and sulfur combine to make 129.0 mg of a compound?

Solution:

1) Determine amount of sulfur:

129.0 mg − 46.4 mg = 82.6 mg

2) Determine millimoles of each element:

(46.4 mg C) (1 mmol C / 12.0 mg C) = 0.00387 mmol C
(82.6 mg S) (1 mmol S / 32.1 mg S) = 0.00257 mmol S

3) Determine lowest whole-number ratio of the mole amounts:

0.00387 mmol C / 0.00257 mmol S = 1.5
0.00257 mmol S / 0.00257 mmol S = 1

4) Empirical formula:

Multiply 1.5 and 1 by 2 to get a whole-number ratio.

C3S2

Wikipoo: "Carbon subsulfide is an inorganic chemical compound with the formula C3S2. This deep red liquid is immiscible with water but soluble in organic solvents. It readily polymerizes at room temperature to form a hard black solid."


Problem #17: Determine the empirical formula of a compound that contains 0.1299 mol C, 0.31176 mol H, and 0.10392 mol O.

Classical method:

1) Divide through by the smallest value:

C ---> 0.1299 mol / 0.10392 mol = 1.25 <--- do NOT round off to 1
H ---> 0.31176 mol / 0.10392 mol = 3
O ---> 0.10392 mol / 0.10392 mol = 1

2) We need to multiply the above values by a factor that creats a full set of whole numbers. The key is the 1.25. Think of it as 5/4. What is the smallest factor that you multiply 5/4 by to get a whole number? The answer is 4 because 5/4 times 4 = 5. So:

C ---> 1.25 x 4 = 5
H ---> 3 x 4 = 12
O ---> 1 x 4 = 4

3) The empirical formula is:

C5H12O4

Alternate method:

The ratio of H : O is 0.31176 : 0.10392 = 3:1. There are 3 times as many H's as O's.

And the ratio of H to C is 0.31176 : 0.1299 = 2.4. If I multiply by 5, I'll get a whole number ratio. 2.4 = 2.4*5/5 = 12/5. So for every 12 H's there are 5 C's.

So one possibility would be 5C's, 12H's, and then 4 O's (1/3 as many as H's).


Problem #18: A compound, a sample of which contains 0.398 moles of the substance , and consists of 38.2 g C, 4.8 g H, and 38.2 g O. Determine its empirical and molecular formulas.

Solution:

1) Determine total mass of compound present:

38.2 + 4.8 + 38.2 = 81.2 g

2) Determine molecular weight of substance:

81.2 g / 0.398 mol = 204 g/mol

3) Let us convert grams to moles:

C ---> 38.2 / 12.01 = 3.1807
H ---> 4.8 / 1.01 = 4.7525
O ---> 38.2 / 16.00 = 2.3875

4) Divide by smallest:

C ---> 3.1807 / 2.3875 = 1.33
H ---> 4.7525 / 2.3875 = 1.99 = 2
O ---> 2.3875 / 2.3875 = 1

5) Think of 1.33 as 4/3. Multiply through by 3 to obtain this:

C4H6O3

This is the empirical formula and its weight is 102

6) Determine molecular formula:

204/102 = 2

The molecular formula is C8H12O6


Problem #19: A 8.059 gram sample of iron is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 10.37 g. Determine the empirical formula of the metal oxide.

Solution:

1) All of the additional mass comes from oxygen, so:

(10.37 g total) − (8.059 g Fe) = 2.311 g O

(8.059 g Fe) / (55.8452 g Fe/mol) = 0.14431 mol Fe

(2.311 g O) / (15.99943 g O/mol) = 0.14444 mol O

2) Divide by the smaller number of moles:

(0.14431 mol Fe) / (0.14431 mol Fe) = 1.000

(0.14444 mol O) / (0.14431 mol Fe) = 1.001

3) The empirical formula:

FeO

Problem #20: The molar mass of Y is 30.97 g/mol and 0.248 mol of the compound contains 117.53 g. What is the most likely identity of compound X3Y2 by formula and name?

Solution:

In one mole of X3Y2, there are two moles of Y

(30.97) (2) = 61.94 g

473.91 − 61.94 = 411.97 g <--- this is the weight of X in one mole of X3Y2

There are three moles of X in one mole of X3Y2

411.97 g / 3 mol = 137.3 g/mol <--- molar mass of X

Look on the periodic table for 137.3 and 30.97

Ba3P2

Barium phosphide


Problem #21: A certain metal oxide has the formula MO where M denotes the metal. A 39.46 g sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, 31.70 g of the metal is left over. Calculate the atomic mass of M and identify the element.

Solution:

Mass O ---> 39.46 − 31.70 = 7.76 g

Moles O ---> 7.76 g / 15.9994 g/mol = 0.485 mol

Because the formula is MO, the moles M also equal 0.485 mol

molar mass = 31.70 g / 0.485 mol = 65.4 g/mol

copper


Problem #22: A 12.370 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide.

Solution:

1) Moles of Mo2O3 present (will be used in step 4):

12.370 g / 239.917 g/mol = 0.0515595 mol

2) Mass of oxygen added to obtain new oxide:

13.197 g − 12.370 g = 0.827 g

3) Convert mass of oxygen added to moles:

0.827 g / 15.9994 g/mol = 0.051689 mol

4) Number of oxygen atoms added per formula unit of Mo2O3:

0.051689 mol / 0.0515595 mol = 1

5) Mo2O3 plus one more O atom becomes Mo2O4. The empirical formula of the new oxide is this:

MoO2

Problem #23: A 18.75 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 22.50 g. Identify the empirical formula of the new oxide.

Solution:

1) Moles of Mo2O3 present:

18.75 g / 239.917 g/mo = 0.078152 mol

2) Moles of Mo in 0.078152 mol of Mo2O3:

(0.078152 mol) (2) = 0.156304 mol Mo

This is becase there are 2 moles of Mo in every mole of Mo2O3.

3) Determine mass of molybdenum present:

(0.156304 mol) (95.96 g/mol) = 14.9989 g

Since only oxygen was added, this is the mass of Mo in the first oxide as well as the second oxide.

4) Determine mass of oxygen in the second oxide:

22.50 g − 14.9989 g = 7.5011 g

5) Determine moles of Mo and O in unknown oxide:

Mo ---> 0.156304 mol (from step 2)
O ---> 7.5011 g / 15.9994 g/mol = 0.468836 mol

6) Divide through by smallest:

Mo ---> 0.156304 mol / 0.156304 mol = 1
O ---> 0.468836 mol / 0.156304 mol = 3

7) The empirical formula is:

MoO3

Ten Examples      Problems #1 - 10      Return to Mole Table of Contents

Determine 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 molecular formula using the Ideal Gas Law

Determine the formula of a hydrate   Hydrate lab calculations