Question

Mass of metal=0.0291. Volume of gas collected over water=25.67mL. Temperature=24.1C. Atmospheric pressure=754.6mmHg. Note: Metal and gas have a 1:1 ratio in the balanced equation. The Formula weight of the metal is? Steps. 1) calculate partial pressure of hydrogen. 2) calculate moles of hydrogen. 3) calculate moles of metal. 4) calculate atomic mass of metal. 5) calculate relative average deviation in ppt

Answer 1

Answers:

1) 732.1 mmHg; 2) 0.001 014 mol; 3) 0.001 014 mol; 4) 28.7 g/mol; 5) 187 ppt.

Explanation:

1) Partial pressure of hydrogen

You are collecting the gas over water, so

$p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}$

$p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}$

$p_{\text{atm}} = \text{754.6 mmHg}$

At 24.1 °C, $p_{\text{H}_{2}\text{O}} = \text{22.5 mmHg}$

$p_{\text{H}_{2}} = \text{754.6 mmHg} - \text{22.5 mmHg} = \textbf{732.1 mmHg}$

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2) Moles of H₂

We can use the Ideal Gas Law.

pV = nRT                Divide both sides by RT and switch

n = (pV)/(RT)

p = 732.1 mmHg                                                 Convert to atmospheres

p = 732.1/760                                                      Do the division

p = 0.9633 atm

V = 25.67 mL                                                        Convert to litres

V = 0.025 67 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 24.1 °C                                                              Convert to kelvins

T = (24.1 + 273.15 ) K = 297.25 K                          Insert the values

n = (0.9633 × 0.025 67)/(0.082 06 × 297.25)     Do the multiplications

n = 0.02473/24.39                                                 Do the division

n = 0.001 014 mol

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3) Moles of metal

The partial chemical equation is

M + … ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.001 014 × 1/1                          Do the operations

Moles of M = 0.001 014 mol M

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4) Atomic mass of M

Atomic mass = mass of M/moles of M     Insert the values

Atomic mass = 0.0291/0.001 014             Do the division

Atomic mass = 28.7 g/mol

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5) Relative deviation in ppt

Your metal must be in Group 2 because of the 1:1 molar ratio of M:H₂.

The metal with the closest atomic mass is Mg (24.305 g/mol).

Relative deviation in ppt = |Experimental value – Theoretical value|/Theoretical value × 1000

Relative deviation = |28.7 – 23.405|/23.405 × 1000     Do the subtraction

Relative deviation = |4.39|/23.405 × 1000                      Do the operations

Relative deviation = 187 ppt