Problems #1 -10 (incomplete)

Ten Examples | Return to Mole Table of Contents |

Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

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

Determine the formula of a hydrate

The example and problem questions only, no solutions

**Problem #1:** An oxybromide compound, KBrO_{x}, where x is unknown, is analyzed and found to contain 52.92% Br. What is the value of x?

**Solution:**

1) The three elements exist in the compound in a fixed ratio:

K : Br : O = 1 : 1 : x

2) Use the atomic weights to write an expression for the percent bromine:

52.92 = (1 times 79.9) / { [(1) (39.1)] + [(1) (79.9)] + [(x) (16.0)] }52.92 = 79.9 / [119 + 16x]

3) We know x must be an integer, so let us try x = 1:

52.92 = 79.9 / [119 + (16) (1)]79.9 / [119 + 16] = 59.185 (which does not equal 52.92)

4) Let us try x = 2:

52.92 = 79.9 / [119 + (16) (2)]79.9 / [119 + 32] = 52.914

4) The formula for KBrO_{x} is:

KBrO_{2}

**Problem #2:** An oxide of an element has the formula M_{2}O_{7} and contains 50.45 % oxygen. (a) What is the gram-atomic weight of the element? (Identify the most probable element for M to be.)

**Solution to (a):**

1) Assume 100 g of the oxide is present. That means 50.45 g of O and 49.55 g of M are present.

2) Determine moles of O:

1 mol O 50.45 g O x ––––––– = 3.153 mol O 16.00 g

3) Determine moles of M:

2 mol M x ––––––– = –––––––––– 7 mole O 3.153 mol O x = 0.90086 mol M

4) Gram-atomic weight of M:

49.55 g / 0.90086 mol = 55.00 g/mol

**Solution to (b):**

On a periodic table, we find that manganese weighs 54.94 g/mol.In addition, Mn

_{2}O_{7}is known to exist.Therefore, we conclude that the unknown element is most probably manganese.

**Problem #3:** Metal, M, forms an oxide, M_{2}O_{3}, which contains 81.9% of the metal by weight. Find the atomic weight of M.

**Solution #1:**

1) Assume 100 g of the compound is present.

That means 81.9 g of the compound is M and 18.1 g is O.

2) Determine moles of O:

18.1 g / 15.9994 g/mol = 1.13129 mol

3) The molar ratio between M and O is 2:3. Use this ratio to determine moles of M:

2 is to 3 as x is to 1.13129x = 0.7541933 mol

4) Determine the atomic weight of M:

81.9 g / 0.7541933 mol = 108.6 g/mol

**Solution #2:**

1) Molar mass of M_{2}O_{3} is:

2(M) + 3(O) <--- where M and O are the atomic weights of element M and oxygenleading to:

2M + 48.0

2) Set up percent M calculation:

2M / (2M + 48.0) = 0.819

3) Solve:

2M = 0.819(2M + 48.0)2M = 1.638 M + 39.312

0.362 M = 39.312

M = 108.6 g/mol

**Problem #4:** An ionic compound formed from aluminum and a group VIA element is 18.55 percent Al by mass. What is the formula of the compound?

**Solution:**

1) The compound will have a generic formula of Al_{2}X_{3}. To solve the problem, we first assume that 100.0 g of the compound is present. 18.55 g of that will be Al. Convert that mass into moles:

18.55 g Al / 26.98 g/mol = 0.6875 mol Al

2) Using the formula of the compound, calculate moles of X:

(0.6875 mol Al) (3 mol X / 2 mol Al) = 1.031 mol X

3) The molar mass of X:

(100.0 − 18.55 g) / 1.031 mol = 79.0.0 g/mol

4) The element in group VIA with a molar mass of 79 g/mol is selenium. So the compound is Al_{2}Se_{3}, named aluminum selenide.

**Problem #5:** An element X forms two oxides whose formulas are XO_{3} and X_{2}O_{3}. One of the oxides contains 52% X by weight and has a formula weight of 99.98 g/mol. What is the formula of this oxide?

(a) CrO_{3}

(b) Al_{2}O_{3}

(c) Mg_{2}O_{3}

(d) VO_{3}

**Solution:**

Assume 100 g (or 99.98 g, if you wish) of the oxide is present. This means X accounts for 52 g and O for 48 g.48 g / 16 g/mol = 3 mol of O present in the 100 g

Therefore, the atomic weight of X will be either 52 (if XO

_{3}works out) or 26 (if X_{2}O_{3}works out)CrO

_{3}---> from the periodic table, the atomic weight of Cr is 52. This is the correct answer. 52 from the Cr and 48 from the O account for the 100 g in the sample.Al

_{2}O_{3}---> The 52 would be divided by 2 to get 26. However, Al is known to weigh 27. This is a wrong answer.Mg

_{2}O_{3}---> The 52 would be divided by 2 to get 26. However, Mg is 24. This is a wrong answer.VO

_{3}---> the atomic weight of V is 51. Not being 52 makes it a wrong answer.

**Problem #6:** A metal (M) forms a compound with the formula MCl_{3}. If the compound contains 65.57% Cl by mass, what is the identity of the metal?

**Solution:**

1) Assume 100 g of the compound is present. This turns our percents into mass. Let us calculate moles:

Cl ---> 65.57 g / 35.543 g/mol = 1.8448 mol

2) From the formula, the molar ratio of M to Cl is 1:3. How many moles of M are present?

1 is to 3 as x is to 1.8448 molx = 0.61493 mol

3) We know that there was 34.43 g of M present in our 100 g of MCl_{3}. We can now determine the atomic weight of M:

34.43 g / 0.61493 mol = 56.0 g/molThis is very close to the atomic weight of iron, which is 55.845 g/mol

Our compound is FeCl

_{3}.

**Problem #7:** A binary compound between an unknown element X and hydrogen contains 92.258% X and 7.742% hydrogen by mass. If the formula of the compound is X_{4}H_{4}, calculate the atomic mass of X and identify what element it is.

**Solution:**

1) Assume 100 g of X_{4}H_{4} is present. This means 92.258 g of X and 7.742 g of H are present.

2) Determine moles of H present:

H ---> 7.742 g / 1.008 g/mol = 7.68056 mol

3) There is a 1:1 molar relationship between X and H. Therefore, 7.68056 mol of X is present. Determine the molar mass of X:

92.258 g / 7.68056 mol = 12.01 g/molCarbon.

**Problem #8:** Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

**Solution:**

1) Assume 100 grams of the compound is present. That means 69.9 g is Fe and 30.1 g is dioxygen. (The fact that it is __ di__oxygen will be important just below.)

2) Determine moles of iron and dioxygen

iron ---> 69.9 / 55.845 = 1.2517 mol of Fe

dioxygen ---> 30.1/31.9988 = 0.94066 mol of dioxygen

3) However, the empirical formula of a chemical compound is the simplest whole-number ratio of __atoms__ present in a compound. Dioxygen is not O, it is O_{2}. Rewrite the mole amounts:

iron ---> 69.9 / 55.845 = 1.2517 mol of Fe

oxygen ---> 30.1/15.9994 = 1.88132 mol of oxygen atoms

4) Another way to think about this:

two atoms of O for every dioxygen (O_{2})therefore:

(0.94066 mol O

_{2}) (2 mol O / 1 mol O_{2}) = 1.88132 mol O

5) Divide through by lowest:

iron ---> 1.2517 1.2517 = 1

oxygen ---> 1.88132 / 1.2517 = 1.5

6) The empirical formula is Fe_{2}O_{3}.

7) This link goes to a discussion of this problem on an "answers" website. Some of the responses miss the point about the use of 'dioxygen' in the question. The writer put it there deliberately to create havoc with the student (as seen in the discussion by the original poster. The idea is to get the student to see that the empirical formula is the smallest whole-number ratio of __atoms__ in the formula, not __molecules__.

**Problem #9:** A compound M_{2}O_{3} contains 68% by mass of element X. Determine what element M is.

**Solution:**

1) Assume 100 g of the compound is present.This means:

68 g of M is present

32 g of O is present

2) Determine moles of oxygen:

32 g / 16 g/mol = 2.0 mol

3) There is a 2:3 molar ratio between M and O. Therefore:

2 is to 3 as x is to 2.0x = 1.3333 mol of M

4) Calculate atomic mass of M and identify the element:

68 g / 1.3333 mol = 51 g/molVanadium

**Problem #10:** Someday!