Question

an organic compound contains 10% of carbon 0.84% of hydrogen and rest is chlorine if it's molicular mass is 49 then what are it's emperical formula and molicular formula

Answer 1

I hope you help !!!!!
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=>. Find chorine’s percentage.
Cl= 100- (10.4+0.84) =100- 11.24= 88.76%Find the mass of teach components in grams:
carbon: 10.4% so 10.4g out of 100g of total mixture.
hydrogen: 0.84g
Chlorine: 88.76gfind the no. of moles of each (no. of moles= total mass/ atomic mass):
Carbon: 10.4/ 12 = 0.8
hydrogen: 0.84/1 = 0.84
Chlorine: 88.76/35= 2.5devide smallest mole vale by all mole values and write their ratio:
1:1:3 = C:H:ClWrite the respective values of ratio as subscrits of the chemical symbol:
C1H1Cl3 = CHCl3 
SO CHCL3 IS THE EMPIRICAL FORMULA OF THE COMPOUND.Find the empirical formula mass:
Mass of C + Mass of H + Mass of 3 atoms of Cl= empirical formula mass
12+1+(3x35)= 118 uDivide the molecular mass (that is given in the question) with the empirical formula mass:
119.5/118 = 1.01 which is approximately 1Multiply all the subscripts of the empirical formula with the answer of the previous step (which is 1 in this case):
C(1x1)H(1x1)CL(3x1)= CHCl3
Hence, the molecular formula of the compound is CHCl3

Answer 2

Answer:

The molecular formula of the compound is $CHCl_{3}$.

Explanation:

To find the percentage of chlorine.

$Cl= 100- (10.4+0.84)$

$=100- 11.24$

$= 88.76$%

To find the mass of each component in grams:

Carbon: $10.4$%

So $10.4g$ out of $100g$ of the total mixture.

Hydrogen: $0.84g$

Chlorine: $88.76g$ find the number of moles of each

Number of moles = total mass/ atomic mass

Carbon: $10.4/ 12 = 0.8$

Hydrogen: $0.84/1 = 0.84$

Chlorine: $88.76/35= 2.5$

Divide the smallest mole value by all mole values and write their ratio:

$1:1:3 = C:H:Cl$

$C_{1} H_{1} Cl_{3} = CHCl_{3}$

So, $CHCl_{3}$ is the empirical formula of the compound.

To find the empirical formula mass:

Mass of C + Mass of H + Mass of 3 atoms of Cl = empirical formula mass

$12+1+(3*35)= 118$

Divide the molecular mass with the empirical formula mass:

$119.5/118 = 1.01...................(1)$

Multiply all the subscripts of the empirical formula to $(1)$, we get

$C_{ (1*1)}H_{ (1*1)}CL_{ (3*1)}= CHCl_{3}$

Hence, the molecular formula of the compound is $CHCl_{3}$.

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