Question

Citric acid is composed of only carbon, hydrogen, and oxygen. when a 0.5000 g sample of citric acid was burned, it produced 0.6871 g of co2 and 0.1874 g of h2o. the molar mass of the compound is 192 g/mol. what are the empirical and molecular formulas of citric acid

Answer 1

The answer is:

C6H8O7

The explanation:

1- we will use the masses provided to get the number of moles and the mass of each atom that was in your starting sample of citric acid:

when All of the Carbon from the citric acid ends up in the CO2, as CO2 is the only product that contains Carbon.

a) now we will get the moles of CO2:

when moles = mass / molar mass

moles CO2 formed = 0.6871 g / 44.01 g/mol

moles CO2 = 0.01561 moles CO2

and when Each CO2 molecule has 1 C atom in it

That means moles of C in CO2 = 0.01561 moles

So, moles C in the 0.5000 g citric acid was 0.01561 moles

- now we need to get mass of C:

mass C = molar mass x moles

mass C = 12.01 g/mol x 0.01561 mol

mass C = 0.18748 g

B) Now for H, all the H from the citric acid ends up in the H2O

we need to get moles H2O and mass:

moles H2O = mass / molar mass

moles H2O = 0.1874 g / 18.016 g/mol

moles H2O = 0.01040 moles H2O

when each H2O molecule has 2 H atoms

Therefore moles H = 2 x 0.01040 = 0.02080 moles H

There were 0.02080 moles H in the 0.5000 g citric acid

-So, mass H = molar mass x moles

mass H = 1.008 g/mol x 0.02080 mol

mass H = 0.02097 g

Now that you know the masses of C and H that were in the citric acid you can determine the mass that must have been O. It is the difference between the mass of citric acid and the masses of C and H.

c- now we can get mass and moles of O:

mass O = mass citric acid - mass C - mass H

mass O = 0.5000 g - 0.18748 g - 0.02097 g

mass O = 0.29155 g

moles O = mass / molar mass

molaes O = 0.29155 g / 16.00 g/mol

moles O = 0.01822 moles O in 0.5000 g citric acid

So now you have worked out the moles of each atom in the citric acid sample. The empirical formula is the lowest whole number ration of atoms in the molecule.

the ratio of moles C : moles H : moles O

is

0.01561 mol : 0.02080 mol : 0.01822 mol

To get this to the lowest whole number ratio divide every number by the smallest number

(0.01561 / 0.01561) : (0.02080 / 0.01561) : (0.01822/ 0.01561)

= 1 : 1.333 : 1.167

You can see this is not a whole number ratio. Multiply the ratio by the smallest number required to produce all whole numbers.

Let's try 2

= 6 : 8 : 7

So empirical formula (smallest whole number ratio of atoms in the molecule) is

C6H8O7

To Find the molecular formula divide the molar mass provided by the molar mass of the empirical formula.

This tells you how many times the empirical formula fits into the molecular formula.

-now we will get molar mass:

molar mass C6H8O7 = (12.01 x 6) + (8 x 1.008) + (7 x 16.00)

= 192 g/mol

molar mass / empirical mass

= 192 g/mol / 192 g/mol

= 1

so, The empirical mass = molar mass

That means the empirical formula is the same as the molecular formula.

Answer 2

Answer:-
`C_6H_8O_7` is empirical formula as well as molecular formula.

Solution:- Mass of carbon dioxide produced = 0.6871 g

Carbon dioxide has only one carbon it means one mole of carbon dioxide has one mole of C. So, let's convert grams of carbon dioxide to moles and then using mole ratio the moles of C as:

`0.6871g CO_2((1mol CO_2)/(44g CO_2))((1mol C)/(1mol CO_2))`

= 0.0156 mol C

Let's do the same for calculating the moles of H from given 0.1874 grams of water. 1 mole of water contains two moles of H, so the ratio is 1:2. The set up would be:

`0.1874g H_2O((1mol H_2O)/(18g H_2O))((2mol H)/(1mol H_2O))`

= 0.0208 mol H

Let's convert moles of C and H to grams and then we subtract the sum of grams of C and H from mass of compound to calculate the grams of O .

Calculations for grams of C:

`0.0156mol C((12g C)/(1molC))`

= 0.1872 g C

Calculations for grams of H:

`0.0208mol H((1g)/(1mol))`

= 0.0208 g H

grams of O = 0.5000 g - (0.1872 g + 0.0208 g)

grams of O = 0.292 g

Let's convert grams of O to moles as:

`0.292g O((1mol)/(16g))`

= 0.01825 mol O

We know that, "An empirical formula is the simplest whole number ratio of all the atoms present in the compound.

To calculate the mole ratio, let's divide the moles of each by the least one of them. Least one is C. So, let's divide the moles of each by the moles of C as:

`C=(0.0156)/(0.0156)` = 1

`H=(0.0208)/(0.0156)` = 1.33

`O=(0.01825)/(0.0156)` = 1.17

The mol ratio is not a whole number ratio. So, let's multiply it by 6 and what we get is:

C = 1*6 = 6

H = 1.33*6 = 8

O = 1.17*6 = 7

So, the empirical formula is:
`C_6H_8O_7` .

Empirical formula mass = 6(12)+8(1)+7(16) = 192

Given molar mass of the compound is also 192. It means, the molecular formula and empirical formula are exactly same here.

Hence, the empirical formula and molecular formula of the compound is
`C_6H_8O_7`.