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ipn [44]
3 years ago
12

The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4% O, and its molar mass is

102 g/mol. Calculate the empirical and molecular formulas for this compound.
Chemistry
2 answers:
weeeeeb [17]3 years ago
8 0
Let our basis for the calculation be 1 mol of the substance.
      (1 mol)(102 g/mol) = 102 g
Determine the amount of C, H, and O in mole.
          C = (102 g)(0.588)(1 mol/12 g) = 4.998 mols C
          H = (102 g)(0.098)(1 mol/1 g) = 9.996 mols H
          O = (102 g)(0.314)(1 mol/16 g) = 2 mols O

The empirical formula of the substance is C5H10O2. The molar mass of the empirical formula is 102. This means that this is also its molecular formula.

Vlada [557]3 years ago
7 0

Answer : The empirical and molecular formula of the compound is, C_5H_{10}O_2

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 58.8 g

Mass of H = 9.8  g

Mass of O = 31.4 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{58.8g}{12g/mole}=4.9moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.8g}{1g/mole}=9.8moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{31.4g}{16g/mole}=1.9moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{4.9}{1.9}=2.5

For H = \frac{9.8}{1.9}=5.16\approx 5

For O = \frac{1.9}{1.9}=1

The ratio of C : H : O = 2.5 : 5 : 1

To make the ratio in a whole number we are multiplying ratio by 2, we get:

The ratio of C : H : O = 5 : 10 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_5H_{10}O_2

The empirical formula weight = 5(12) + 10(1) + 2(16) = 102 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{102}{102}=1

Molecular formula = (C_5H_{10}O_2)_n=(C_5H_{10}O_2)_1=C_5H_{10}O_2

Therefore, the molecular formula of the compound is, C_5H_{10}O_2

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