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8 votes
8 votes
When copper is heated with an excess of sulfur, copper(l) sulfide is

formed. In a given experiment, 0.0970 moles of copper was heated
with excess sulfur to yield 5.59 g copper(1) sulfide. What is the
percent yield?

User Josh Black
by
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1 Answer

15 votes
15 votes

Answer:

72.4%

Step-by-step explanation:

Step 1: Write the balanced equation

2 Cu + S ⇒ Cu₂S

Step 2: Calculate the theoretical yield, in moles, from 0.0970 moles of Cu

The molar ratio of Cu to Cu₂S is 2:1.

0.0970 mol Cu × 1 mol Cu₂S/2 mol Cu = 0.0485 mol Cu₂S

Step 3, Convert the theoretical yield to mass

The molar mass of Cu₂S is 159.16 g/mol.

0.0485 mol × 159.16 g/mol = 7.72 g

Step 4: Calculate the percent yield

We will use the following expression.

%yield = experimental yield/theoretical yield × 100%

%yield = 5.59 g/7.72 g × 100% = 72.4%

User Jaksky
by
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