Problems #26-50

All Examples & Problems (no solutions)

**Problem #26:** A 0.204 gram sample of a metal, M, reacts completely with sulfuric acid according to:

M + H_{2}SO_{4}---> MSO_{4}+ H_{2}

A volume of 213 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 torr and the temperature is 25.0 °C. Calculate the molar mass of the metal.

The vapor pressure of water at various temperatures can be found in this table.

**Solution:**

1) Use Dalton's Law of Partial Pressures to determine the pressure of the dry H_{2}:

756.0 - 23.8 = 732.2 torr

2) Converting that to atm gives:

732.2 torr / 760.0 torr/atm = 0.9634 atm

3) Use the ideal gas law:

PV = n RT(0.9634 atm) (0.213 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.0083915 mol H

_{2}

4) From the balanced equation, 1 mole of H_{2} is formed when 1 mol of metal M is reacted. Therefore, the moles of metal that reacted was:

0.0083915 mol M

5) The molar mass of M is:

0.204 g / 0.0083915 mol = 24.3 g/molThe metal was magnesium.

**Problem #27:** A common way to obtain a pure metal from its impure metal oxide is to react the oxide with carbon, expressed generically as:

2MO(s) + C(s) ---> 2M(s) + CO_{2}(g)

If 5.00 g of an unknown metal oxide (MO) reacted with excess carbon and formed 738 mL of CO_{2} at 200.0 °C and 0.978 atm, what is the identity of the metal?

**Solution:**

1) Using PV = nRT, solve for moles of CO_{2}:

(0.978 atm) (0.738 L) = (n) (0.08206 L atm / mol K) (473 K)n = 0.0186 mol

2) Determine moles of oxygen atoms in metal oxide:

For every one mole of CO_{2}produced, two moles of MO were consumed.1 is to 2 as 0.0186 is to x

x = 0.0372 mol

3) Determine molar mass of MO:

5.00 g / 0.0372 mol = 134.44 mol

4) Determine atomic weight of M:

x + 16.00 = 134.44x = 118.44 g/mol

M is tin.

**Problem #28:** A compound of P and F was analyzed as follows: heating 0.2324 g of the compound in a 378 cm^{3} flask turned all of it to gas, which had a pressure of 97.3 mmHg at 77 °C. Then, the gas was mixed with calcium chloride solution which turned all of the F to 0.2631 g of CaF_{2}. Determine the molecular formula of the compound.

**Solution:**

1) Molecular weight of the gas:

(97.3 / 760) (0.378) = (n) (0.08206) (350)n = 0.001684967 mol

0.2324 g / 0.001684967 mol = 138 g/mol

We will use this to move from the empirical formula to the molecular formula.

2) Mass of F:

0.2631 g times (38.0 g / 78.074 g) = 0.128055 g(38.0 g / 78.074 g) is called a gravimetric factor. It is the decimal percent of F in CaF

_{2}.

3) Moles of F:

0.128055 g / 19.0 g/mol = 0.00674 mol

4) Moles of P:

0.2324 g minus 0.128055 g = 0.104345 g0.104345 g / 31.0 g/mol = 0.003366 mol

5) Empirical formula:

P ---> 0.003366 / 0.003366 = 1

F ---> 0.00674 / 0.003366 = 2PF

_{2}

6) Molecular formula:

PF_{2}weighs 31 + 38 = 69138 / 69 = 2

PF

_{2}times 2 = P_{2}F_{4}

**Problem #29:** A metal chloride reacts with silver nitrate solution to give a precipitate of silver chloride according to following equation:

MCl_{2}+ 2AgNO_{3}---> M(NO_{3})_{2}+ 2AgCl

When a solution containing 0.4750 g of metal chloride is made to react with silver nitrate, 1.435 grams of silver chloride are formed. Identify the metal.

**Solution:**

1) Determine moles of AgCl produced:

1.435 g AgCl / 143.32 g/mol = 1.001256 x 10^{-2}mol AgCl

2) Determine moles of MCl_{2} consumed:

1.001256 x 10^{-2}mol AgCl x (1 mol MCl_{2}/ 2 mol AgCl) = 5.00628 x 10^{-3}mol MCl_{2}

3) Determine molar mass of MCl_{2}:

0.4750 g / 5.00628 x 10^{-3}mol = 94.88 g/mol

4) Determine the atomic weight and identity of M:

The atomic of M will be:94.88 minus the mass of 2 chlorine atoms:

94.88 minus (2 x 35.453) = 23.97 g/molM is magnesium

**Problem #30:** An unidentified metal M reacts with an unidentified halogen X to form a compound MX_{2}. When heated the compound decomposes by the reaction:

2MX_{2}(s) ---> 2MX (s) + X_{2}(g)

When 1.12 g of MX_{2} is heated, 0.720 g of MX is obtained along with 56.0 mL of X_{2} gas (at STP).

a) What is the atomic mass and the identity of the halogen X?

b) What is the atomic mass and identity of the metal M?

**Solution:**

1) Calculate moles of X_{2} collected:

PV = nRT(1.00 atm) (0.0560 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.00250 mol

2) Calculate mass of X_{2} collected:

1.12 g minus 0.720 g = 0.40 g

3) Calculate molecular weight of X_{2} and determine identity of X

0.40 g/0.00250 moles = 160 g/molhalf of 160 is 80, the atomic weight of X

bromine

4) Determine moles of MX:

MX : X_{2}molar ratio is 2 : 1therefore, 0.0050 moles of MX

5) Calculate molar mass of MX:

0.720 g/0.0050 moles = 144 g/mole

6) Calculate atomic weight of M:

144 g/mol minus 80 g/mol = 64 g/molecopper

**Problem #31: **A metal sulfate has the formula M_{2}SO_{4}. 10.99 g of the compound was dissolved in water to make 500.0 cm^{3} of solution. A 25.0 cm^{3} sample was removed and reacted with an excess of BaCl_{2}(aq) to produce a precipitate of BaSO_{4}, which when dried had a mass of 1.167 g.

a) Determine the number of moles of BaSO_{4}precipitated.

b) Determine the concentration of M_{2}SO_{4}

c) Identify M

**Solution:**

1) Moles of BaSO_{4} precipitated from 25 cm^{3}:

1.167 g / 233.4 g/mol = 0.005 mol

2) Moles of BaSO_{4} that would have precipitated from 500 cm^{3}:

0.005 mol is to 25 as x is to 500x = 0.100 mol of BaSO

_{4}

3) The molarity of the M_{2}SO_{4} is this:

0.100 mol / 0.500 L = 0.200 MThe 0.100 mole comes from the fact that there is a 1:1 molar ratio between M

_{2}SO_{4}and BaSO_{4}M

_{2}SO_{4}+ CaCl_{2}---> BaSO_{4}+ 2MCl

4) What is M?

0.1 mol of sulfate weighs 96.061 x 0.1 = 9.6061 g10.99 minus 9.6061 = 1.3839 g of M

^{+}in solutionThere were 0.2 mol of M

^{+}in the 500 cm^{3}[from this: M_{2}SO_{4}(s) ---> 2M^{+}(aq) + SO_{4}^{2-}(aq)]1.3839 g / 0.2 mol = 6.9 g/mol

Li

_{2}SO_{4}

**Problem #32:** An element, X, forms two compounds with bromine: XBr_{2} and XBr_{4}. When 10.00 grams of the XBr_{2} is reacted with excess bromine, 14.35 g of XBr_{4} is formed. Identify X.

**Solution:**

14.35 - 10.00 = 4.35 g of Br_{2}reacted4.35 g / 159.808 g/mol = 0.02722 mol of Br

_{2}XBr

_{2}+ Br_{2}---> XBr_{4}XBr

_{2}and Br_{2}react in a 1:1 molar ratioTherefore, 10.00 g represents 0.02722 mol of XBr

_{2}10.00 g / 0.02722 mol = 367 g/mol

In 367.377 g of XBr

_{2}, there is 159.808 g of bromine367.377 - 159.808 = 207.569 <--- this is the atomic weight of X

X is lead.

**Problem #33:** Exactly 4.32 g of oxygen gas was required to completely combust a 2.16 g sample of a mixture of methanol and ethanol:

(1) How many moles of ethanol are contained within the sample?

(2) What is the percentage by weight of methanol in the sample?

**Solution:**

1) Balanced chemical equations:

CH_{3}OH + 1.5O_{2}---> CO_{2}+ 2H_{2}O

C_{2}H_{5}OH + 3O_{2}---> 2CO_{2}+ 3H_{2}O

2) Let x = mass of methanol; let y = mass of ethanol

x + y = 2.16[(1.5) (x/32)] + [(3) (y/46)] = 4.32/32

3) Using x = 2.16 - y:

[(1.5) ((2.16-y)/32)] + [(3) (y/46)] = 0.135[(3.24 - 1.5y) / 32] + [3y / 46] = 0.135

4) Multiply each side by 32, then by 46:

(46) (3.24 - 1.5y) + (32) (3y) = 198.72149.04 - 69y + 96y = 198.72

27y = 49.68

y = 1.84 g

x = 0.32 g

5) Answers to (1) and (2):

moles ethanol ---> 1.84 g / 46 g/mol = 0.040 molpercent methanol ---> (0.32 g / 2.16 g) * 100 = 14.8%

**Problem #34:** A 3.41 g sample of a metallic element, M, reacts completely with 0.0158 mol of a gas, X_{2}, to form 4.53 g MX. What are the identities of M and X?

**Solution:**

2M + X_{2}---> 2MXWe know that M and X

_{2}react in a 2:1 molar ratio.Therefore, 0.0158 mol of X

_{2}reacts with twice that many moles of M, 0.0316 molWe can now determine the atomic weight of M:

3.41 g / 0.0316 mol = 107.9 g/mol

From the periodic table, we see that M is silver.

4.53 - 3.41 = 1.12 g <--- this is the mass of X in MX

We know that there is a 1:1 molar ratio between M and MX, therefore 0.0316 of MX was produced. Since we know the formula is MX, we know that 0.0316 mol of X is involved.

1.12 g / 0.0316 mol = 35.44 g/mol

This atomic weight is within experimental error for X to be chlorine.

**Problem #35:** When 2.3 moles of X reacts with 1.6 moles of Y, 71 grams of Z are produced. What is the molar mass of Z?

3X + 4Y ---> 5Z

This reaction has a 50% yield.

**Solution:**

1) Determine the limiting reagent:

X: 2.3 / 3 = 0.77

Y: 1.6 / 4 = 0.4Y is the limiting reagent.

2) How much Z is produced at 100% yield:

4 is to 5 as 1.6 is to xx = 2.0 mol

3) The reaction has only 50% yield, so 1.0 mole of Z was produced. The molar mass of Z is:

71 g / 1.0 mol = 71 g/mol

**Problem #36:** Someday!