Answer:
Empirical formula is C₅H₆O₄₀₇.
Explanation:
Given data:
Mass of carbondioxide = 152.5 mg
Mass of water vapors = 36.67 mg
Mass of toluene = 45.62 mg
Empirical formula = ?
Solution:
First of all we will convert the mg into g.
Mass of carbondioxide = 152.5/1000 = 0.1525 g
Mass of water vapors = 36.67/ 1000 = 0.0367 g
Mass of toluene = 45.62 /1000 =0.0456 g
Percentage of each atoms:
C = 0.1525/0.0456 × 12/ 44 = 0.91
H = 0.0367/0.0456 × 2/ 18 = 0.089
O = 100 - ( 0.91+0.089)
O = 100- 0.999 = 99
Number of gram atoms:
Number of gram atoms of C = O.91/12 = 0.076
Number of gram atoms of H = 0.089/1 = 0.089
Number of gram atoms of O = 99 / 16 = 6.188
Atomic ratio:
C : H : O
0.076/0.076 : 0.089/0.076 : 6.188/0.076
1 : 1.17 : 81.4
C : H : O = 5 ( 1: 1.7 : 81.4)
C : H : O = 5 : 6 : 407
Empirical formula is C₅H₆O₄₀₇.
The empirical formula of toluene is C8H1
The empirical formula is the simplest formula of a compound which shows the ratio of the atoms present.
Mass of carbon = 152.5 * 10^-3g/44 g/mol * 12 g = 0.04 g
Number of moles of carbon = 0.04 g/12 g/mol = 0.033 moles
Mass of hydrogen = 35.67 * 10^-3/18 g/mol * 2 g = 0.00396 g
Number of moles of hydrogen = 0.00396 g/ 1 g/mol = 0.00396 moles
Divide through by the lowest number of moles;
C - 0.033 moles/ 0.00396 moles, H - 0.00396 moles/ 0.00396 molesC C- 8 H - 1
The empirical formula of toluene is C8H1
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