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Acompound is used as a food additive.The compound has a molar mass of 176.124grams/mole A692.6-gram sample undergoes decomposition, producing 283.4 grams of carbon, 31.7 grams of hydrogen, and 3774 grams of oxygen What is the molecular formula of the compound? A. C_(3)H_(4)O_(3) B. C_(4)H_(8)O_(4) C. C_(6)H_(6)O_(6) D. C_(6)H_(8)O_(6)

Question

Acompound is used as a food additive.The compound has a molar mass of 176.124grams/mole A692.6-gram sample undergoes decomposition,
producing 283.4 grams of carbon, 31.7 grams of hydrogen, and 3774 grams of oxygen What is the molecular formula of the compound?
A. C_(3)H_(4)O_(3)
B. C_(4)H_(8)O_(4)
C. C_(6)H_(6)O_(6)
D. C_(6)H_(8)O_(6)

Acompound is used as a food additive.The compound has a molar mass of 176.124grams/mole A692.6-gram sample undergoes decomposition, producing 283.4 grams of carbon, 31.7 grams of hydrogen, and 3774 grams of oxygen What is the molecular formula of the compound? A. C_(3)H_(4)O_(3) B. C_(4)H_(8)O_(4) C. C_(6)H_(6)O_(6) D. C_(6)H_(8)O_(6)

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EllisProfessional · Tutor for 6 years

Answer

D. \( C_{6}H_{8}O_{6} \)

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## Step 1: <br />First, we determine the number of moles of each element in the sample:<br />- Carbon: \( \frac{283.4 \text{ grams}}{12.01 \text{ grams/mole}} = 23.6 \text{ moles} \)<br />- Hydrogen: \( \frac{31.7 \text{ grams}}{1.008 \text{ grams/mole}} = 31.4 \text{ moles} \)<br />- Oxygen: \( \frac{377.4 \text{ grams}}{16.00 \text{ grams/mole}} = 23.6 \text{ moles} \)<br /><br />## Step 2: <br />To find the empirical formula, we divide each mole value by the smallest mole value (23.6 moles) to get the simplest whole number ratio:<br />- Carbon: \( \frac{23.6}{23.6} = 1 \)<br />- Hydrogen: \( \frac{31.4}{23.6} \approx 1.33 \) (rounded to 1)<br />- Oxygen: \( \frac{23.6}{23.6} = 1 \)<br /><br />## Step 3: <br />The empirical formula is thus CHO.<br /><br />## Step 4: <br />The molar mass of the empirical formula is \( 12.01 + 1.008 + 16.00 = 29.018 \text{ grams/mole} \).<br /><br />## Step 5: <br />To find the molecular formula, we divide the given molar mass (176.124 grams/mole) by the molar mass of the empirical formula (29.018 grams/mole) to get a factor of approximately 6.<br /><br />## Step 6: <br />Multiplying each subscript in the empirical formula by this factor, we get the molecular formula \( C_{6}H_{8}O_{6} \).
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