Problems #11 - 20

Ten Examples | Problems #1 - 10 | 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 a percent composition

Determine the formula of a hydrate

Only the examples and problems, no solutions

**Problem #11:** 11.2 g of a metal carbonate, containing an unknown metal, M, was heated to give the metal oxide and 4.40 g CO_{2}. What is the identity of the metal M?

**Solution:**

4.40 g CO_{2}x (1 mole CO_{2}/ 44.0 g) = 0.10 moles CO_{2}MCO

_{3}produces CO_{2}in a 1:1 molar ratio.0.10 moles CO

_{2}x (1 mole MCO_{3}/ 1 mole CO_{2}) = 0.10 mole MCO_{3}11.2 grams MCO

_{3}/ 0.10 moles = 112 g / mole (this is the molar mass of MCO_{3})Formula mass of just CO

_{3}= (1) (12) + (3) (16) = 60Atomic mass of M ---> 112 − 60 = 52

Cr has an atomic mass of 51.99.

**Problem #12:** 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. Given that O has an atomic mass of 16.00 amu, 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

**Problem #13:** A 30.6 g sample of the compound X_{2}O_{3} contains 14.4 g of oxygen atoms. What is the atomic mass of element X?

**Solution:**

1) Determine moles of O present:

14.4 g / 16.00 g/mol = 0.900 mol

2) Use the X:O molar ratio of 2:3 to determine moles of X in the compound:

2 is to 3 as y is to 0.900y = 0.600 mol

3) Determine the atomic mass (in g/mol) of element X:

30.6 g − 14.4 g = 16.2 g16.2 g / 0.600 mol = 27 g/mol

Although not asked for, element X is aluminum.

**Problem #14:** Given a solution of XCl_{2} has a molar concentration of 0.0250 mol dm¯^{3} and a mass concentration of 2.375 g dm¯^{3}, find the relative atomic mass of X and identify its most probable identity.

**Solution:**

1) Assume 1 dm^{3} is present. That means 2.375 g (from the mass conc.) is the same amount as 0.0250 mol (from the molar conc.) of XCl_{2}. How many grams is in one mole?

2.375 is to 0.0250 as x is to 1x = 95 g/mol

2) In XCl_{2}, the Cl_{2} portion accounts for 71 g/mol. The remainder is X.

95 − 71 = 24 g/mol <---that's the answerThe most probable identity of X is magnesium. The substance is MgCl

_{2}, magnesium chloride.

**Problem #15:** In an experiment, hydrogen gas was passed over the metal oxide, MO_{x}. It was found that the mass of the oxide was reduced from 1.59 g to 1.27 g. (a) Calculate the percentage composition of metal M and oxygen in the metal oxide. (b) Then, find the value of x if the molar mass of the oxide is 79.5 g/mol and identify element M.

**Solution:**

1) Masses of each element:

M ---> 1.27 gO ---> 1.59 − 1.27 = 0.32 g

2) Percent composition:

M ---> 1.27 / 1.59 = 0.79874 = 79.9%O ---> 1 − 0.79874 = 0.20126 = 20.1%

3) In 79.5 g of the compound (note that this is one mole of the compound), 79.9% of it is M:

(79.5 g) (0.799) = 63.5205 g79.5 − 63.5205 = 15.9795 g

3) Moles of oxygen that are in one mole of the compound:

15.9795 g / 15.9994 g/mol = 1.00 molThe value for x is 1

4) The mass of M tells us it is copper and MO_{x} is CuO, copper(II) oxide.

**Problem #16** A chloride of a metal M contains 65.5% chlorine. The mass of 0.100 L vapour of metal chloride at STP is 0.72 g. What is the empirical formula of the metal chloride? Identify what element M most probably is.

**Solution:**

1) Calculate molecular weight of the compound:

PV = nRT(1.00 atm) (0.100 L) = (n) (0.08206) (273 K)

n = 0.0044638 mol

0.72 g / 0.0044638 mol = 161.3 g/mol

2) Total mass of Cl in one mole of the metal chloride:

(161.3 g) (0.655) = 105.65 g

3) This is how many Cl are in the formula:

105.65 / 35.5 = 3

4) This is the atomic weight of the metal:

161.3 − 105.65 = 55.65 (the atomic weight of iron)empirical formula ---> FeCl

_{3}

**Problem #17:** 31.87 g of a solid compound, known to have the formula X_{3}N_{2}, is produced when 25.85 g of element X is reacted with excess nitrogen. What is the molar mass of the element and what is the name of the element?

**Solution:**

1) Determine how much nitrogen reacted:

31.87 − 25.85 = 6.02 g

2) Determine how many moles of nitrogen 6.02 grams is:

6.02 g / 14.007 g/mol = 0.4297851 mol

3) In the formula X_{3}N_{2}, there is a 3:2 molar ratio between X and N. Determine how many moles of X are present:

3 is to 2 as y is to 0.4297851y = 0.64467765 mol

4) Determine the molar mass of X:

25.85 g / 0.64467765 mol = 40.1 g/molX is calcium.

**Problem #18:** 1.250 g of a metal, M, is reacted with excess sulfuric acid, yielding 1.860 g of a compound whose molecular formula is MSO_{4}. Determine the identity of the element M.

**Solution:**

1) Determine the amount of sulfate in the compound:

1.860 − 1.250 = 0.610 g

2) Determine moles of sulfate (SO_{4}^{2}¯) present:

0.610 g / 96.061 g/mol = 0.006350132 mol

3) The molar ratio between M and sulfate is 1:1. Therefore:

0.006350132 mol of M is present

4) Determine the atomic mass of M:

1.250 g / 0.006350132 mol = 196.846 g/molThe closest element to our value (with an atomic mass of 196.96657 g/mol) is gold.

5) This is a problem where rounding off too much could generate a different answer. If you had used 0.0064 mol, you would had obtained an answer of 195.3125 g/mol. This answer could lead you to identifying platinum (with an atomic mass of 195.084 g/mol) as the answer.

**Problem #19:** A metal oxide of formula M_{x}O is heated until it completely decomposes into the pure metal and oxygen gas. it was found that 11.71 g of oxygen was produced when 69.00 g of the oxide decomposed. What is the identity of M?

**Solution:**

1) Determine mass of M after decomposition is complete:

69.00 − 11.71 = 57.29 g

2) Determine moles of oxygen gas produced:

11.71 g / 31.9988 g/mol = 0.365951 mol O_{2}

3) Determines moles of atoms of O:

(0.365951 mol O_{2}) (2 mol O / 1 mol O_{2}) = 0.731902 mol OThis is done because whole-number ratios in empirical formulas are ratios of atoms, not of molecules.

4) Let us assume 'x' is 1, leading to a formula of MO. This means, then, that 0.731902 mol of M is present. Determine the atomic mass of M, based on a formula of MO:

57.29 g / 0.731902 mol = 78.276 g/molSelenium has a mass of 78.79 g/mol

5) Le us assume 'x' = 2, giving a formula of M_{2}O. This means 1.463804 mol of M is present when 0.731902 mol of O is present. Determine the atomic mass of M based on a formula of M_{2}O:

57.29 g / 1.463804 mol = 39.14 g/molPotassium has an atomic mass of 39.098 g/mol

6) Potassium is a better candidate, but selenium is plausible. Different tests (such as density, reaction with water) would be needed to make a final determination.

**Problem #20:** A metal X forms two different chlorides. 12.7 g of chloride A contain 7.10 g and 16.3 g of chloride B contains 10.7 g of chlorine. Determine the formula of the compound.

**Solution:**

1) Determine moles of chloride present in A and B:

moles Cl in A ---> (7.10 g Cl) (1 mol / 35.453 g) = 0.2003 mol

moles Cl in B ---> (10.7 g Cl) (1 mol / 35.453 g) = 0.3018 mol

2) Determine the smallest whole-number ratio between the two chlorine amounts:

Cl in A = 0.2003 / 0.2003 = 1

Cl in B = 0.3018 / 0.2003 = 1.51 = 1.5Multiply the 1 to 1.5 ratio by two to obtain the smallest whole-number ratio of Cl in A to Cl in B as 2 to 3.

3) Determine the mass of X in A and B:

mass X in A ---> 12.7 − 7.10 = 5.6 g

mass of X in B ---> 16.3 − 10.7 = 5.6 g

4) In both A and B, there are equal masses, therefore equal number of moles of X. This allows us to determine a partial formula for compounds A and B:

X_{z}Cl_{2}

X_{z}Cl_{3}

5) Let us speculate about z by considering compound A:

0.2 moles of Cl are presentAssume a 1:2 ratio of X to Cl <--- For compound B, it would be to assume a 1:3 ratio

0.1 mol of X is present

5.6 g / 0.1 mol = 56 g/mol

A reasonable conclusion about X is that it is iron.

6) With z = 1, the two formulas would be:

FeCl_{2}

FeCl_{3}

7) Here is another solution to this problem. It includes discussion about the consequences of assuming a 1:1 ratio between X and Cl. You might also be interested in this solution.

8) In my notes, I also found a solution that utilizes the Law of Multiple Proportions.

mass of element 1 (compound B) ––––––––––––––––––––––––––– mass of element 2 (compound B) –––––––––––––––––––––––––––––––––– = a small number ratio mass of element 1 (compound A) ––––––––––––––––––––––––––– mass of element 2 (compound A)

9) Substituting:

10.7 g ––––– 5.6 g 1.5 –––––––––––––– = –––––– 7.1 g 1 ––––– 5.6

10) Here is a statement of the Law of Multiple Proportions:

"When two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers"

11) From the analysis in step 9, we can conclude the following:

For every 1 Cl in compound A, there are 1.5 Cl in compound B. Therefore, based upon Dalton's atomic theory, there are two Cl atoms in compound A for every three Cl atoms in compound B.

Ten Examples | Problems #1 - 10 | 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 a percent composition