The Atomic Weight of a Metal Data ntHai volume hnai velunC 1. Mass of Magnesium

Question

The Atomic Weight of a Metal Data ntHai volume hnai velunC 1. Mass of Magnesium (g) 2. Volume of Hydrogen Gas Collected (mL) 3. Temperature (C) 4. Temperature (K) (show calculation) 5. Barometric Pressure (mm Hg) Trial 1 o.012 l.d ml 21.8 c Trial 2 2t.. 294.8 295.1 n Height Difference between Water Levels (mm H,O Scola ure-1S Mercury Equivalent of Height Difference ms.sty dies dele Water Difference (show calculation) 13.6 39.2s 1%. Vapor Pressure of Water (will be given--mm Hg) Pressure of Dry Hydrogen Gas 18.05 zs.80 -1e.OS Barometric Pressure - Mercury Equivalent y of Height Difference Vapor Pressure of Tu. 29 Water (show calculation) Number of Mole of Hydrogen Gas Collected Number of Mole of Metal Molar Mass of a Solid (g/mol) Average Molar Mass 6 Percent of Error Lab Report: The Atomic Weight of a Metal

Explanation / Answer

The balanced chemical reaction-

Mg + 2HCl = MgCl2 + H2

Trial 1

Given:

Mass of Mg = 0.012g

Volume of hydrogen gas collected = 11.4 ml = 0.0114 L

Temperature = 294.8 K

Atmospheric pressure = 756.5mm Hg

Height of water column = 349.25mm H2O

Mercury equivalent of height difference = 349.25/13.6 = 25.68 mmHg

Determine the pressure of hydrogen gas in mmHg

P dry H2 = P total – P water vapor

PH2 = (756.5 mmHg – 25.68 mmHg) = 730.82mmHg.

PH2 = 730.82 mmHg

Calculate the number of moles H2 gas produced.

We can calculate the number of moles H2 gas produced, by using equation-

PV = nRT

R is gas constant = 62.363 L mmHg K1 mol1

n = PV/RT

n = (730.82 mmHg x 0.0114 L)/(62.363 L mmHg K1 mol1 x 294.8 K)

n = 0.000543 moles

Calculate the number of moles metal reacted in experiment

According to the balanced reaction, 1 moles of metal produces 1moles of hydrogen gas

So moles of metal = H2 = 0.000543 moles

Molar mass of metal = Mass of metal/ moles of metal

Molar mass of metal = 0.012/0.000543 = 22.1 g/mol

Trial 2

Given:

Mass of Mg = 0.010g

Volume of hydrogen gas collected = 11.5 ml = 0.0115 L

Temperature = 295.1 K

Atmospheric pressure = 756.5mm Hg

Height of water column = 361.95mm H2O

Mercury equivalent of height difference = 361.95/13.6 = 26.61 mmHg

Determine the pressure of hydrogen gas in mmHg-

PH2 = (756.5 mmHg -26.61 mmHg ) = 729.89mmHg

Calculate the number of moles H2 gas produced.

n = PV/RT

n = (729.89 mmHg x 0.0115 L)/(62.363 L mmHg K1 mol1 x 295.1 K)

n = 0.000456 moles

Calculate the number of moles metal reacted in experiment

According to the balanced reaction, 1 moles of metal produces 1moles of hydrogen gas

So moles of metal = H2 = 0.000456 moles

Molar mass of metal = Mass of metal/ moles of metal

Molar mass of metal = 0.010/0.000456 = 21.93 g/mol

Average molar mass of metal =22.1 g/mol +21.93 g/mol/2 = 22.01 g/mol

Percent error =[ (24.305 g/mol -22.01 g/mol)/ 24.305 g/mol] x 100 = 9.4%

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