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katrin2010 [14]
3 years ago
13

When we write the formula for an ionic compound, we are merely indicate the "molecules" in the compound with that formula. expla

in?
Chemistry
1 answer:
belka [17]3 years ago
3 0
When writing an ionic compound formula, a "molecular" form is used.  The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻  ⇒ Al₂(SO₄)₃
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A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.
Alexxandr [17]

<u>Answer:</u> The empirical and molecular formula of the compound is C_4H_4O and C_8H_8O_2 respectively

<u>Explanation:</u>

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = \frac{0.313}{0.078}=4.01\approx 4

For Hydrogen  = \frac{0.316}{0.078}=4.05\approx 4

For Oxygen  = \frac{0.078}{0.078}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is C_4H_4O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

n=\frac{130g/mol}{68g/mol}=1.9\approx 2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2

Hence, the empirical and molecular formula of the compound is C_4H_4O and C_8H_8O_2 respectively

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Explanation:

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