A compound is used as a food additive. The compound has a molar mass of 176.124 grams/mole. A 692.5-gram sample undergoes decomposition, producing 283.4 grams of carbon, 31.7 grams of hydrogen, and 377.4 grams of oxygen. What is the molecular formula of the compound?
A. $C_3H_4O_3$
B. $C_4H_6O_4$
C. $C_6H_6O_6$
D. $C_6H_8O_6$
Answer (2)
Calculate the moles of each element using the given masses and molar masses.
Determine the mole ratio of C:H:O and find the empirical formula C 3 H 4 O 3 .
Calculate the molar mass of the empirical formula and divide the compound's molar mass by it to find the multiplier.
Multiply the subscripts in the empirical formula by the multiplier to obtain the molecular formula: C 6 H 8 O 6 .
Explanation
Problem Analysis We are given the mass of each element in a sample of a compound, and we need to determine the molecular formula of the compound. We will first calculate the number of moles of each element, then find the empirical formula, and finally determine the molecular formula using the given molar mass.
Moles Calculation First, we need to find the number of moles of each element. The molar masses are: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.008 g/mol, Oxygen (O) = 16.00 g/mol. We have 283.4 grams of Carbon, 31.7 grams of Hydrogen, and 377.4 grams of Oxygen.
Calculating Moles of Carbon Now, let's calculate the moles of each element:
Moles of Carbon = 12.01 g/mol 283.4 g = 23.596 mol
Calculating Moles of Hydrogen Moles of Hydrogen = 1.008 g/mol 31.7 g = 31.45 mol
Calculating Moles of Oxygen Moles of Oxygen = 16.00 g/mol 377.4 g = 23.5875 mol
Finding Mole Ratio Next, we find the mole ratio by dividing each mole value by the smallest mole value, which is approximately 23.59.
Calculating the Ratio C : 23.5875 23.596 ≈ 1.00036 ≈ 1 H : 23.5875 31.45 ≈ 1.333 ≈ 3 4 O : 23.5875 23.5875 = 1
Empirical Formula To get whole numbers, we multiply the ratio by 3: C : H : O = 3 : 4 : 3 So, the empirical formula is C 3 H 4 O 3 .
Molar Mass of Empirical Formula Now, we calculate the molar mass of the empirical formula C 3 H 4 O 3 :
(3 * 12.01) + (4 * 1.008) + (3 * 16.00) = 36.03 + 4.032 + 48.00 = 88.062 g/mol
Finding the Multiplier We divide the molar mass of the compound (176.124 g/mol) by the molar mass of the empirical formula (88.062 g/mol): 88.062 176.124 = 2.00 ≈ 2
Molecular Formula Finally, we multiply the subscripts in the empirical formula by 2 to get the molecular formula: C ( 3 ∗ 2 ) H ( 4 ∗ 2 ) O ( 3 ∗ 2 ) = C 6 H 8 O 6
Final Answer Therefore, the molecular formula of the compound is C 6 H 8 O 6 .
Examples
Understanding molecular formulas is crucial in various fields. For instance, in pharmacology, knowing the precise molecular formula of a drug helps determine its interactions with the body. Imagine a scientist developing a new vitamin supplement. By accurately determining the molecular formula, they can ensure the correct dosage and efficacy. If the supplement's molecular formula is C 6 H 8 O 6 , it indicates the exact number of carbon, hydrogen, and oxygen atoms in each molecule, which is vital for its intended function and safety.
The molecular formula of the compound is C 6 H 8 O 6 , derived from calculating moles of each element and finding the empirical formula C 3 H 4 O 3 , then using the given molar mass to determine the final formula. The chosen multiple choice option is D. C 6 H 8 O 6 .
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