Gravimetric Analysis - GRE Subject Test: Chemistry

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Question

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . If a vinegar sample contains $0.1024\ moles$ of acetic acid, calculate the percent (mass/mass) of the $CH_{3}COOH$ in the vinegar sample.

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Answer

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5%\ CH_{3}COOH$

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Question

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . If a vinegar sample contains $0.1024\ moles$ of acetic acid, calculate the percent (mass/mass) of the $CH_{3}COOH$ in the vinegar sample.

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Answer

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5%\ CH_{3}COOH$

Compare your answer with the correct one above

Question

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . If a vinegar sample contains $0.1024\ moles$ of acetic acid, calculate the percent (mass/mass) of the $CH_{3}COOH$ in the vinegar sample.

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Answer

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5%\ CH_{3}COOH$

Compare your answer with the correct one above

Question

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . If a vinegar sample contains $0.1024\ moles$ of acetic acid, calculate the percent (mass/mass) of the $CH_{3}COOH$ in the vinegar sample.

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Answer

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5%\ CH_{3}COOH$

Compare your answer with the correct one above

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