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GRE Subject Test: Chemistry : Other Statistics

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #11 : Analytical Chemistry

Determine the percentage yield if the expected yield of $ZnSO_{4}$ in the above reaction is 40.911 grams but the actual yield is 35.322 grams.

Possible Answers:

$86.3 \%$

$81\%$

$3.8\%$

$2.0\%$

Correct answer:

$86.3 \%$

Explanation:

In chemistry, percentage yield is a term used to quantify the efficiency of a chemical reaction. In order to calculate this quantity, the value for the actual yield and theoretical yield is needed. The actual yield is the amount of product obtained during a chemical reaction. The theoretical yield is the maximum amount of product possible to be obtain by a chemical reaction which is based on the amount of reactants used. Below is the formula needed to calculate the percentage yield.

$Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$Percentage\ Yield =\frac{35.322g}{40.911g}\times100=86.3 \%$

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Example Question #1 : Other Statistics

0.0031 moles of NaOH was reacted with 0.0071 moles of HCl giving an actual yield of 0.1733 grams of NaCl . What is the percent yield of this reaction?

Possible Answers:

$22.9 \%$

$61.5 \%$

$89.7 \%$

$95.7 \%$

Correct answer:

$95.7 \%$

Explanation:

HCl is the limiting reactant because based on the equation, HCl and NaOH react in a 1:1 mole ratio. However, there is 0.0040 moles more of HCl than NaOH present. Therefore the amount of NaCl produced is limited by the amount of NaOH present in the reaction. Based on the equation, for every 1 mole of NaOH reacted, 1 mole of NaCl was produced:

$moles\ of\ NaOH\ reacted=moles\ of\ NaCl\ produced$

Therefore,

$0.0031\ moles\ NaCl\times \frac{58.44g}{moles}=0.1812g\ NaCl$

$Percent\ Yield =\frac{actual\ yield}{expected\ yield}\times100$

$Percent\ Yield =\frac{0.1733g}{0.1812g}\times100=95.7 \%$

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Example Question #12 : Analytical Chemistry

Determine the percentage yield if the expected yield of $BaSO_{4}$ in the above reaction is 1.2124 grams but the actual yield is 0.9749 grams.

Possible Answers:

$12.7 \%$

$40.2 \%$

$65.9 \%$

$80.4 \%$

Correct answer:

$80.4 \%$

Explanation:

$Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$Percentage\ Yield =\frac{0.9749g}{1.2124g}\times100=80.4 \%$

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All GRE Subject Test: Chemistry Resources

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GRE Subject Test: Chemistry : Other Statistics

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #11 : Analytical Chemistry

Example Question #1 : Other Statistics

Example Question #12 : Analytical Chemistry

All GRE Subject Test: Chemistry Resources

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