5. Calculate the molar ratio between copper and oxygen. . According to the mola

Question

5. Calculate the molar ratio between copper and oxygen. . According to the mola oxide? Conclusions 1. How does the empirical formula calculated from experiment 2 compare to the empirical formula calculated in experiment 1? 2. Similar to copper, magnesium also reacts with the oxygen in the air when heated to formm magnesium oxide. Given the experimental data in the table below, what is the empirical formula of magnesium oxide? The molar mass of magnesium is 24.31 g/mol and the molar mass of oxygen is 16.00 g/mol. mass of empty crucible mass of crucible with magnesium before reaction mass of crucible after heating 64.000g 65.125 g 65.866 g 3. The reported molar mass for the copper oxide formed in experiment 1 is 79.55 g/mol. What is the molecular formula of the oxide formed in experiment 1? The molar mass of copper is 63.55 g/mol and the molar mass of oxygen is 16.00 g/mol.

Explanation / Answer

Data and Observations:

Mass of crucible = 88.000 g

Mass of crucible + copper (before burning) = 98.000 g

Mass of 10g Cu is heated = 100.518 g

Mass of cu = 98.000 g - 88.000 g = 10.000 g

Mass of product = 100.518 g - 88.000 g = 12.518 g

10 grams of copper becomes 12.518 grams of copper oxide,

10 grams of copper reacts with 2.518 grams of oxygen.

Convert 10 grams of copper and 2.518 grams of oxygen to their respective moles

10 g Cu x (1 mol Cu / 63.55 g Cu) = 0.157 mol Cu

2.518 g O x (1 mol O / 16.00 g O) = 0.157 mol O

Find the lowest whole number ratio between them: 0.157/0.157 = 1

Mole ratio of copper and oxygen is 1:1, so the empirical formula is CuO

2) Mass of crucible = 64.000 g

Mass of crucible + magnesium (before burning) = 65.125 g

Mass of crucible after heating = 65.866 g

Mass of Mg = 65.125 g - 64.000 g = 1.125 g

Mass of product = 65.866 g - 64.000 g = 1.866 g

1.125 grams of Mg becomes 1.866 grams of magnesium oxide, meaning it absorbs 0.741 grams of oxygen.

Convert 1.125 grams of Mg and 0.741 grams of oxygen to their respective moles by dividing with their molar mass

Molar mass of Mg = 24.31 g/mol

Molar mass of oxygen = 16.0 g/mol

Moles of Mg = 1.125 g/24.31 g/mol = 0.0463 moles

Moles of oxygen = 0.741g /16.0 g/mol = 0.0463 moles

Find the lowest whole number ratio between them: 0.0463/0.0463 = 1

Mole ratio of magnesium and oxygen is 1:1, so the empirical formula is MgO

3) Reported molar mass of copper oxide formed in experiment 1 = 79.55 g/mol

Molecular formula of oxide formed

Molar mass of copper = 63.55 g/mol

Molar mass of oxygen = 16.0 g/mol

Cu + xO ---> CuxO

63.55 g/mol + x (16.0 g/mol) = 79.55 g/mol

x (16.0 g/mol) = 79.55 g/mol - 63.55 g/mol

x (16.0 g/mol) = 16.0 g/mol

x = 1

Molecular formula of oxide formed = CuO

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