Question

The percent composition by mass of certain compound is 35.5 percent carbon, 4.8 percent hydrogen, 37.9 percent oxygen, 8.3 percent nitrogen, and 13.5 percent sodium. What is the molecular formula if the molar mass is roughly 170 grams per mole

Answer 1

Answer : The molecular formula of a compound is,
`C_5H_8O_4NNa`

Solution :

If percentage are given then we are taking total mass is 100 grams.

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

Mass of C = 35.5 g

Mass of H = 4.8 g

Mass of O = 37.9 g

Mass of N = 8.3 g

Mass of Na = 13.5 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Molar mass of N = 14 g/mole

Molar mass of Na = 23 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (35.5g)/(12g/mole)=2.96moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (4.8g)/(1g/mole)=4.8moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (37.9g)/(16g/mole)=2.37moles`

Moles of N =
`\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (8.3g)/(14g/mole)=0.593moles`

Moles of Na =
`\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= (13.5g)/(23g/mole)=0.587moles`

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

For C =
`(2.96)/(0.587)=5.0\approx 5`

For H =
`(4.8)/(0.587)=8.1\approx 8`

For O =
`(2.37)/(0.587)=4.0\approx 4`

For N =
`(0.593)/(0.587)=1.0\approx 1`

For Na =
`(0.587)/(0.587)=1`

The ratio of C : H : O : N : Na = 5 : 8 : 4 : 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`C_5H_8O_4N_1Na_1=C_5H_8O_4NNa`

The empirical formula weight = 5(12) + 8(1) + 4(16) + 1(14) + 1(23) = 169 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

`n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}`

`n=(170)/(169)=1`

Molecular formula =
`(C_5H_8O_4NNa)_n=(C_5H_8O_4NNa)_1=C_5H_8O_4NNa`

Therefore, the molecular of the compound is,
`C_5H_8O_4NNa`

Answer 2

`C_(5)H_(8)NNaO_(4)`

Step-by-step explanation:

`Moles =\frac {Given\ mass}{Molar\ mass}`

% of C = 35.5

Molar mass of C = 12.0107 g/mol

% moles of C =
`(35.5)/(12.0107)` = 2.95569

% of H = 4.8

Molar mass of H = 1.00784 g/mol

% moles of H =
`(4.8)/(1.00784)` = 4.76266

% of N = 8.3

Molar mass of N = 14.0067 g/mol

% moles of N =
`(8.3)/(14.0067)` = 0.59257

% of Na = 13.5

Molar mass of Na = 22.989769 g/mol

% moles of N =
`(13.5)/(22.989769)` = 0.58721

% of O = 37.9

Molar mass of O = 15.999 g/mol

% moles of O =
`(37.9)/(15.999)` = 2.36889

Taking the simplest ratio for C, H, O, N, Na and O as:

2.95569 : 4.76266 : 0.59257 : 0.58721 : 2.36889

= 5 : 8 : 1 : 1 : 4

The empirical formula is =
`C_(5)H_(8)NNaO_(4)`

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12*5 + 1*8 + 14 + 23 + 16*4 = 169 g/mol

Molar mass = 170 g/mol

So,

Molecular mass = n × Empirical mass

170 = n × 169

⇒ n ≅ 1

The molecular formula =
`C_(5)H_(8)NNaO_(4)`