Question

Consider a compound that is 31.17% C, 6.54% H, and 62.29% O by mass. Assume that we have a 100 g sample of this compound. Also consider that the molecular formula mass of this compound is 231.2 amu. What are the subscripts in the actual molecular formula for this compound?

Answer 1

To determine the subscripts in the molecular formula, find the empirical formula and use the molecular formula mass to calculate the subscripts. The empirical formula is C10H22O5 and the molecular formula is C23H50O11.

Step-by-step explanation:

To determine the subscripts in the molecular formula, we need to find the empirical formula first. Given that the compound is 31.17% C, 6.54% H, and 62.29% O by mass, we can assume that we have 31.17 g of C, 6.54 g of H, and 62.29 g of O in a 100 g sample. Convert these mass percentages to moles by dividing each element's mass by its molar mass. The molar mass of C is 12.01 g/mol, H is 1.01 g/mol, and O is 16.00 g/mol. After finding the moles of each element, divide each value by the smallest number of moles to obtain a whole number ratio. The empirical formula of the compound is C10H22O5.

Next, we need to find the molecular formula using the molecular formula mass and the empirical formula. Divide the molecular formula mass (231.2 amu) by the empirical formula mass (102.19 amu) to get a multiplication factor of around 2.26 (rounded to the nearest whole number). Multiply the subscripts in the empirical formula by this factor to obtain the subscripts in the molecular formula. The molecular formula for the compound is approximately C23H50O11.

Answer 2

Answer: The molecular formula will be
`C_6H_(15)O_9`

Step-by-step explanation:

If percentage are given and we are given total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 31.17 g

Mass of H = 6.54 g

Mass of O = 62.29 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (31.17g)/(12g/mole)=2.6moles`

Moles of H=
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (6.54g)/(1g/mole)=6.54moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (62.29g)/(16g/mole)=3.9moles`

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

For C=
`(2.6)/(2.6)=1`

For H =
`(6.54)/(2.6)=2.5`

For O =
`(3.9)/(2.6)=1.5`

The ratio of C: H: O= 1: 2.5: 1.5

Converting them into whole number ratios by multiplying by 2.

Hence the empirical formula is
`C_2H_5O_3`

The empirical weight of
`C_2H_5O_3` = 2(12)+5(1)+3(16)= 77 amu

The molecular weight = 231.2 amu

Now we have to calculate the molecular formula.

`n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=(231.2)/(77)=3`

The molecular formula will be=
`3* C_2H_5O_3=C_6H_(15)O_9`

The molecular formula will be
`C_6H_(15)O_9`