Question

An unknown sugar is found to have a molar mass of 180.18 g/mol. The sugar contains: 40 grams of C, 6.7 grams of H, and 53.3 grams of O.what isthe molecular formula?

Answer 1

The molecular formula is `"C"_6"H"_12"O"_6"`.

This is the molecular formula for 6-carbon monosaccharides, such as glucose, fructose, and galactose.

Explanation:

We first need to find the empirical formula, which represents the smallest whole number ratio of elements in the compound. After we determine the empirical formula, we determine the empirical molar mass.

If the empirical molar mass is the same as the molecular molar mass, then the empirical formula is also the molecular formula.

If the empirical molar mass is not the same as the molecular molar mass, we divide the molecular molar mass by the empirical molar mass, which will give us the factor to multiply by the subscripts in the empirical formula. This will give the molecular formula.

Empirical Formula

Since the masses add to 100, we can start by determining the moles for each element.

Moles of each element

To determine the moles of each element, divide the given mass for of element by its molar mass to get moles. The molar mass of an element is its atomic weight on the periodic table in g/mol.

`"Moles C":` `(40color(red)cancel(color(black)("g C")))/(12.011color(red)cancel(color(black)("g"))/"mol")="3.3 mol"`

`"Moles H":` `(6.7color(red)cancel(color(black)("g H")))/(1.008color(red)cancel(color(black)("g"))/"mol")="6.6 mol"`

`"Moles O":` `(53.3color(red)cancel(color(black)("g O")))/(15.999color(red)cancel(color(black)("g"))/"mol")="3.33 mol"`

Now we need to determine the mole ratio for each element.

Mole ratios: subscripts of the empirical formula

Divide the number of moles for each element by the least number of moles.

`"C":` `(3.3)/(3.3)=1.0`

`"H":` `(6.6)/(3.3)=2.0`

`"O":` `(3.33)/(3.3)=1.0`

The empirical formula is `"CH"_2"O"`.

The empirical molar mass is:

`(1xx"12.011 g/mol C")+(2xx"1.008 g/mol H")+(1xx"15.999 g/mol O")="30.026 g/mol"`

Molecular Formula

Divide the molecular molar mass by the empirical molar mass.

`(180.18color(red)cancel(color(black)("g"))/color(red)cancel(color(black)("mol")))/(30.026color(red)cancel(color(black)("g"))/color(red)cancel(color(black)("mol")))=6.0008~~6"`

Multiply the subscripts in the empirical formula by `6` to get the subscripts of the molecular formula.

`"C"_6"H"_12"O"_6"`

This is the molecular formula for 6-carbon monosaccharides, such as glucose, fructose, and galactose.

Answer 2

We have here...`C_6H_12O_6`...

Explanation:

We interrogate the molar quantities of each element...(note that the elements have been given as percentages as well as masses),

`"Moles of carbon"=(40.0*g)/(12.01*g*mol^-1)=3.33*mol.`

`"Moles of hydrogen"=(6.7*g)/(1.00794*g*mol^-1)=6.65*mol.`

`"Moles of oxygen"=(53.3*g)/(16.00*g*mol^-1)=3.33*mol.`

We divide thru by the SMALLEST molar quantity to get the `"empirical formula.."` `=C_((3.33*mol)/(3.33*mol))H_((6.65*mol)/(3.33*mol))O_((3.33*mol)/(3.33*mol))`...`-=CH_2O`

But the molecular formula is always a whole number multiple of the empirical formula...

`"molecular formula"-={"empirical formula"}_n`..

We solve for `n`..

`nxx{12.011+2xx1.00794+16.00}*g*mol^-1=180.18*g*mol^-1`..

`n=(180.18*g*mol^-1)/({12.011+2xx1.00794+16.00}*g*mol^-1)`

`n=6`

And `"molecular formula"=C_6H_12O_6`..