Find the empirical formula of the compound which contains 33.18 percent of carbon 4.60 percent of hydrogen 29.49 percent of Oxygen and 32.72 percentage of chlorine respectively.
Answers
Mark Me BRAINLIEST
Explanation:
The empirical formula of a compound is, C_3H_5O_2ClC
3
H
5
O
2
Cl
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 = 33.18 g
Mass of H = 4.60 g
Mass of O = 29.49 g
Mass of Cl = 32.72 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Molar mass of Cl = 35.5 g/mole
Step 1 : convert given masses into moles.
Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{33.18g}{12g/mole}=2.765moles
molar mass of C
given mass of C
=
12g/mole
33.18g
=2.765moles
Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.60g}{1g/mole}=4.60moles
molar mass of H
given mass of H
=
1g/mole
4.60g
=4.60moles
Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{29.49g}{16g/mole}=1.843moles
molar mass of O
given mass of O
=
16g/mole
29.49g
=1.843moles
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{32.72g}{35.5g/mole}=0.9217moles
molar mass of Cl
given mass of Cl
=
35.5g/mole
32.72g
=0.9217moles
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = \frac{2.765}{0.9217}=2.99\approx 3
0.9217
2.765
=2.99≈3
For H = \frac{4.60}{0.9217}=4.99\approx 5
0.9217
4.60
=4.99≈5
For O = \frac{1.843}{0.9217}=1.99\approx 2
0.9217
1.843
=1.99≈2
For Cl = \frac{0.9217}{0.9217}=1
0.9217
0.9217
=1
The ratio of C : H : O : Cl = 3 : 5 : 2 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = C_3H_5O_2Cl_1=C_3H_5O_2ClC
3
H
5
O
2
Cl
1
=C
3
H
5
O
2
Cl
Therefore, the empirical of the compound is, C_3H_5O_2ClC
3
H
5
O
2
Cl