# Question c2dd9

Mar 28, 2017

$\text{CH"_ 2"O}$

#### Explanation:

The first thing you need to do here is to figure out how many moles of carbon were present in the original carbohydrate.

To do that, use the molar mass of carbon

2.48 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "0.2065 moles C"

Now, you know that mass must be conserved, which means that the mass of carbon and the mass of water must be equal to the mass of the carbohydrate.

In other words, you can say that dry heating the carbohydrate drove off

overbrace("6.2 g")^(color(blue)("the mass of the carbohydrate")) = overbrace("2.48 g")^(color(purple)("the mass of carbon")) + overbrace(m_"water")^(color(darkorange)("the mass of water"))

Therefore, you will have

${m}_{\text{water" = "6.2 g " - " 2.48 g}}$

${m}_{\text{water" = "3.72 g}}$

Use the molar mass of water to convert this to moles

3.72 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.2065 moles H"_2"O"

As you can see, the sample of carbohydrate contained equal numbers of moles of carbon and of water, which can only mean that carbon and water are present in a $1 : 1$ mole ratio in the compound.

 "C"/("H"_ 2"O") = (0.2065 color(red)(cancel(color(black)("moles"))))/(0.2065 color(red)(cancel(color(black)("moles")))) = 1/1

This represents the smallest whole number ratio that exists between carbon and water in the carbohydrate.

Therefore, you can say that for

"C" _ x("H"_ 2"O")_ y#

you have

$x = y = 1$

and

$\text{C"_ 1 ("H"_ 2"O")_ 1 implies "CH"_2"O} \to$ the empirical formula

Mar 28, 2017

On heating the compound ${C}_{x} {\left({H}_{2} O\right)}_{y}$ decompses into Cabon and water as follows

${C}_{x} {\left({H}_{2} O\right)}_{y} \to x C + y {H}_{2} O$

From this equation the ratio of masses Cabon and water is

${m}_{C} / {m}_{{H}_{2} O} = \frac{12 x}{18 y} = \frac{2 x}{3 y}$

where molar mass of

$C \to 12 g \text{/} m o l$

${H}_{2} O \to 18 g \text{/} m o l$

The ratio obtained above is a constant and is independent of the mass of the compound decomposed.

Now it is given that $6.2$ g of the compound on decomposition produces $2.48$ g Carbon and the rest mass should be mass of water produced.

Hence mass of water produced is

$6.2 - 2.48 = 3.72$ g

So

${m}_{C} / {m}_{{H}_{2} O} = \frac{2 x}{3 y} = \frac{2.48}{3.72}$

$\implies \frac{x}{y} = {\cancel{2.48}}^{1.24} / {\cancel{3.72}}^{1.24} \times \frac{\cancel{3}}{\cancel{2}} = \frac{1}{1}$

So the empirical formula of the compound is $C {H}_{2} O$