Question

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

moles of oxoacid equals 44.0 g. For this compound, determine the empirical and molecular formula.

Answer 1

Answer: The molecular formula will be
`C_6H_6O_6`

Step-by-step explanation:

Mass of
`CO_2` = 17.95 g

Mass of
`H_2O`= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =
`(12)/(44)* 17.95=4.89g` of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =
`(2)/(18)* 4.87=0.541g` of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (4.89g)/(12g/mole)=0.407moles`

Moles of H=
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (0.541g)/(1g/mole)=0.541moles`

Moles of O=
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (6.57g)/(16g/mole)=0.410moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(0.407)/(0.407)=1`

For H =
`(0.541)/(0.407)=1`

For O =
`(0.410)/(0.407)=1`

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is
`CHO`.

Hence the empirical formula is
`CHO`

The empirical weight of
`CHO` = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of =
`(44.0)/(0.25)* 1=176g`

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

`n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=(176g)/(29g)=6`

The molecular formula will be=
`6* CHO=C_6H_6O_6`