Problems #1 - 10

Ten Examples | Problems #11 - 20 | Return to Mole Table of Contents |

Determine empirical formula when given mass data

Determine empirical formula when given percent composition data

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

Determine molecular formula using the Ideal Gas Law

Determine the formula of a hydrate Hydrate lab calculations

The example and problem questions only, no solutions

**Problem #1:** An unknown metal M reacts with oxygen to give the metal oxide MO_{2}. Identify the metal based on the following information.

mass of metal: 0.356 g

mass of metal oxide: 0.452 g

**Solution:**

1) Mass of the oxygen in MO_{2}:

0.452 − 0.356 = 0.096 g

2) Moles of oxygen in MO_{2}:

0.096 g / 16.00 g/mol = 0.0060 mol

3) The moles of M in the sample of MO_{2}:

the molar ratio of M and O in the formula is 1 to 21 is to 2 as x is to 0.0060

x = 0.0030 mol

4) Calculate atomic weight and identify M:

0.356 g / 0.0030 mol = 118.7 g/molTin

**Problem #2:** 16.5 g of element X reacts completely with 9.6 g of oxygen to produce a pure sample of XO_{2}. Find the atomic weight and identity of X.

**Solution:**

9.6 g / 16.00 g/mol = 0.60 molmole ratio of X and O in the formula is 1 to 2. Therefore:

1 is to 2 as x is to 0.60

x = 0.30 mol (this is moles of X in the 26.1 g of XO

_{2})16.5 g / 0.30 mol = 55 g/mol

Manganese

**Problem #3:** An element reacts with bromine to give the bromide, MBr_{5}. If 2.009 g of the element gives 10.648 g of MBr_{5}, what is the element?

**Solution:**

1) How much Br is present in the sample?

10.648 − 2.009 = 8.639 g

2) How many moles is this?

8.639 g / 79.904 g/mol = 0.108117 mol

3) From the formula, we know that M and Br are in a 1 to 5 molar ratio. We use a ratio and proportion to get the moles of M.

1 is to 5 as x is to 0.108117x = 0.0216234 mol

4) We now know a mass of M and how many moles that mass is. To get the molar mass of M, we do this:

2.009 g / 0.0216234 mol = 92.9 g/molNiobium

**Problem #4:** A student places 5.00 g of an unknown metal (X) ribbon in a crucible. The crucible is heated until the unknown metal reacted with oxygen to form a white product with the formula X_{2}O_{3} . The mass of product is determined to be 7.19 g.

(a) What is the molar mass of the unknown metal (X)?

(b) What element is the unknown metal?

**Solution:**

1) Mass of oxygen:

7.19 − 5.00 = 2.19 g

2) Moles of oxygen:

2.19 g / 16.00 g/mol = 0.136875 mol

3) Use ratio and proportion to determine moles of X:

2 is to 3 as y is to 0.136875y = 0.09125 mol

4) Determine molar mass of X:

5.00 g / 0.09125 mol = 54.8 g/molManganese

**Problem #5:** A 30.6-g sample of the compound M_{2}O_{3} contains 9.79 g of oxygen atoms. What is the molar mass of element M? Identify the element M most probably is.

**Solution:**

1) Set up the following ratio and proportion:

30.6 9.79 –––––– = –––– 2x + 48 48 9.79 g / 48 g ---> The moles of O in 30.6 g of M

_{2}O_{3}48 g ---> There are 48 g in three moles of oxygen. The three comes from oxygen's subscript in the formula. (I suppose I should write 48 g/3mol.

2x + 48 ---> The molar mas of M

_{2}O_{3}. x is the atomic weight of M, note that is multiplied by two because of the subscripted two in the formula.

2) Cross multiply and divide:

(9.79) (2x + 48) = (48) (30.6)(9.79) (2x + 48) = 1468.8

2x + 48 = 150

2x = 102

x = 51 g/mol

Vanadium

**Problem #6:** From a specific heat measurement, the approximate atomic weight of a metal (M) is found to be 135 Daltons. A 0.2341 g sample of M is heated to constant weight in air to convert it to the oxide The weight of the residue is 0 2745 g. Find the true atomic weight of the metal (and therefore its identity), and determine the formula of the metal oxide.

**Solution:**

1) Determine the moles of M and O present:

M ---> 2341 g / 135 g/mol = 17.34 mol

O ---> 404 g / 16.0 g/mol = 25.25 mol

2) Determine lowest whole-number ratio:

M ---> 17.34 mol / 17.34 mol = 1

O ---> 25.25 mol /17.34 mol = 1.4563) This yields a ratio of 2 to 2.9 leading to the conclusion that the empirical formula is M

_{2}O_{3}.4) Note that I scaled the weight values to 2341 g and 2745 g. I did this for convenience. Also, the 404 comes from 2745 minus 2341.

5) Determine the percent composition of M in M_{2}O_{3}:

2341 g / 2745 g = 0.8528 (leave it as a decimal value)

6) The calculation for the percent composition of M in M_{2}O_{3} is this:

2X ––––––– = 0.8528 2X + 48 Where X is the atomic weight of M and 48 is the weight of three O.

7) Solve for X:

2X = 1.7056X + 40.93440.2944X = 40.9344

X = 139

8) This experiment yields lanthanum as the most probable answer. Since our starting data was approximate, more tests would be required. For example, Ba is a possibility, but Ba forms BaO when heated in air. In like fashion, Cs can be excluded based in its formation of Cs_{2}O. However, Ce_{2}O_{3} can form, therefore, if this was a real-world situation, additional tests would be performed to distinguish between La and Ce.

**Problem #7:** Identify M in the compound M_{2}(C_{2}O_{4})_{3} if the mass of the M atoms is 1371 g in a sample containing 5.92 x 10^{24} molecules of M_{2}(C_{2}O_{4})_{3}

**Solution:**

5.92 x 10^{24}molecules of M_{2}(C_{2}O_{4})_{3}contain 2 x 5.92 x 10^{24}= 1.184 x 10^{25}atoms of MMoles of M = (1.184 x 10

^{25}atoms) / (6.022 x 10^{23}atoms/mol) = 19.66 molMolar mass = 1371 g / 19.66 mol = 69.74 g/mol

Gallium (molar mass of 69.723 g/mol) is the most reasonable answer.

**Problem #8:** If 3.72 g of element X exactly reacts with 4.80 g of oxygen to form a compound whose molecular formula is shown, from other experiments to be X_{4}O_{10}, what is the relative atomic mass of X? What element is X?

**Solution:**

1) Let's assume "4.80 g of oxygen" means O:

(4.80 grams O) (1 mol O / 16.0 g O) = 0.30 mol of O(0.30 mol O) (4 mol X / 10 mol O) = 0.12 mol X

3.72 g X / 0.12 mol X = 31.0 g/mol

phosphorous

2) Let's assume "4.80 g of oxygen" means O_{2}:

(4.80 grams O_{2}) (1 mol O_{2}/ 32.0 g O_{2}) (2 mol O / 1 mol O_{2}) = 0.30 mol of O(0.30 mol O) (4 mol X / 10 mol O) = 0.12 mol X

3.72 g X / 0.12 mol X = 31.0 g/mol

phosphorous

**Problem #9:** Determine the atomic weight of element A, given that 8.00 g of A_{2}O_{3} contains 2.40 g of oxygen.

**Solution:**

1) Determine moles of oxygen:

2.40 g / 16.0 g/mol = 0.15 mol

2) Determine moles of A:

2 is to 3 as x is to 0.15x = 0.10 mol

3) Determine atomic weight of A:

8.00 g − 2.40 g = 5.60 g5.60 g / 0.10 mol = 56 g/mol

Although not asked for, element A is most likely iron.

**Problem #10:** A 2.00 g sample of a metal X (that is known to form X^{2+} ions) was added to 0.100 L of 0.500 M H_{2}SO_{4}. After all the metal had reacted, the remaining acid required 0.0637 L of 0.500 M NaOH solution for neutralization. Calculate the atomic weight of the metal and identify the element.

**Solution:**

1) Determine moles NaOH:

(0.500 mol/L) (0.0637 L) = 0.03185 mol

2) Two NaOH required for one H_{2}SO_{4}:

2NaOH + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}O0.03185 mol / 2 = 0.015925 mol of H

_{2}SO_{4}remaining in solution

3) Moles of H_{2}SO_{4} at start:

(0.500 mol/L) (0.100 L) = 0.0500 mol

4) Moles H_{2}SO_{4} used to react with X^{2+}:

0.0500 mol − 0.015925 mol = 0.034075 mol

5) X^{2+} and sulfate react in a 1:1 ratio, so:

0.034075 mol of X^{2+}reacted

6) atomic weight of X^{2+}:

2.00 g / 0.034075 mol = 58.69 g/molNickel

Ten Examples | Problems #11 - 20 | Return to Mole Table of Contents |

Determine empirical formula when given mass data

Determine empirical formula when given percent composition data

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