Question

a compound of only one carbon and chlorine contains 10.1 5% carbon and 89.8 5% chlorine is each molecule weighs 39.3 into 10 raise to power minus 23 grams of molar formula of compound would be​

Answer 1

Answer:

We know that, Molar Mass of a gas = 2 x Vapour density of the gas

Molar mass can be found as:

Mass of one mole of gas = Molar mass of the gas

1 mole of gas at STP = 22.4 L.

5.6 L of gas = 8 g

22.4 L o f gas = (22.4x8) / 5.6 = 32 g

Thus, Mass of one mole of gas = 32 g.

So, molar mass of the gas = 32 g/mol

Vapour density = Molar mass / 2 = 32/2 = 16

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Explanation:

Answer 2

Complete Question:

A compound composed of carbon, hydrogen, and chlorine contains C=10.04% and Cl=89.12%. The vapor density of the compound is 59.75. determine its molecular formula.

Answer:

The molecular formula = $CHCl_{3}$

Explanation:

C= 10.04% and Cl= 89,12

H = 100-(10.04+89.12) = 0.84%

H = 0.84%

Formula:

A number of moles of each element in the compound = Given mass/At. Mass

Calculation:

No. of moles in C = $\frac{10.04}{12} = 0.84$

No. of moles in Cl = $\frac{89.12}{35.5} = 25$

No. of moles in H = 0.84

Therefore, Ratio = C:H:Cl = 0.84: 0.84 : 2.5

C:H:Cl= 1 : 1 : 3 (Approx)

Empirical formula = $C_{1} H_{1}Cl_{3}$

= CHCl$_{3}$

V.D = M/2 or 59,75 = M/2

or M = 59.75 x 2 = 119,5

Molecular formula = (empirical formula)$_{n}$

Thus, molecular mass = n x empirical formula mass

119.5 = n x (12+1+35.5*3) = n(119.5)

or n = 1

The molecular formula = $CHCl_{3}$

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