Problems #11 - 20

Ten Examples | Problems #1 - 10 | 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

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 (H_{2}O_{2}) 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 (H

_{2}O_{2}) has an empirical formula of HO. Water (H_{2}O 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:

WCl_{6}

Comment #1: it occured to me that the writer of the problem could have mistakenly used 71.0 (for Cl_{2}) 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 mol0.027 mol / 0.00903 mol = 3

WCl

_{3}

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 3/3Multiply by 3 to get a 4 to 3 ratio

4) Empirical formula:

P_{4}Se_{3}

I found a mention of P_{4}Se_{3} in this article from 1959.

**Problem #15:** 0.0833 mol of carbohydrate of empirical formula CH_{2}O contain 1.00 g of hydrogen. What is the molecular formula of the carbohydrate?

**Solution:**

1.00 g of hydrogen is 0.500 mol0.0833 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 CH

_{2}O must be scaled up to contain 6.00 moles of H in one mole of the compound.CH

_{2}O times 3 = C_{3}H_{6}O_{3}

**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.C

_{3}S_{2}Wikipoo: "Carbon subsulfide is an inorganic chemical compound with the formula C

_{3}S_{2}. 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:

C_{5}H_{12}O_{4}

**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:

C_{4}H_{6}O_{3}This is the empirical formula and its weight is 102

6) Determine molecular formula:

204/102 = 2The molecular formula is C

_{8}H_{12}O_{6}

**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 X_{3}Y_{2} by formula and name?

**Solution:**

In one mole of X_{3}Y_{2}, 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 X

_{3}Y_{2}There are three moles of X in one mole of X

_{3}Y_{2}411.97 g / 3 mol = 137.3 g/mol <--- molar mass of X

Look on the periodic table for 137.3 and 30.97

Ba

_{3}P_{2}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 gMoles O ---> 7.76 g / 15.9994 g/mol = 0.485 mol

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

MM = 31.70 g / 0.485 mol = 65.4 g/mol

copper

Ten Examples | Problems #1 - 10 | 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