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3 years ago

What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen?

Chemistry

1 answer:

rusak2 [61]3 years ago

8 0

Answer:

The empirical formula is C3H3O

Explanation:

Step 1: Data given

Suppose the mass of the molecule = 100 grams

The molecule contains:

65.5 % Carbon = 65.5 grams

5.5 % Hydrogen = 5.5 grams

29.0% Oxygen = 29.0 grams

Molar mass of C = 12 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 65.5 grams / 12 g/mol = 5.46 moles

Moles H = 5.5 / 1.01 g/mol = 5.45 moles

Moles O = 29.0 grams / 16 g/mol = 1.8125 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 5.46 / 1.8125 = 3

H = 5.45 / 1.8125 = 3

O = 1.8125/1.8125 = 1

The empirical formula is C3H3O

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3 years ago

Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to: