Respuesta :
C – 19,9%, H – 2,2%, N – 11,6%, O – x%
[tex]M=120\frac{g}{mol}[/tex]
1 percentage
The entire molecule is 100% and all the components of the compound add up to 100%.
100% - 19,9% - 2,5% - 11,6% = 66,1%
The compound contains 66,1% oxygen.
2 molar masses
[tex]M_{C}=12,01\frac{g}{mol}[/tex]
[tex]M_{H}=1,008\frac{g}{mol}[/tex]
[tex]M_{O}=15,999\frac{g}{mol}[/tex]
[tex]M_{N}=14,007\frac{g}{mol}[/tex]
3 masses
The compound has a molar mass of 120g/mol. So one molecule weighs 120 g. To find out how much the percentage of a component weighs, you have to calculate it using the molar mass.
carbon
19,8% of 120g
[tex]m=120g*0,198\\m=23,76g[/tex]
One molecule contains 23,76g of carbon.
hydrogen
2,5% of 120g
[tex]m=120g*0,025\\m=3g[/tex]
One molecule contains 3g of hydrogen.
oxygen
66,1% of 120g
[tex]m=120g*0,661\\m=79,32g[/tex]
One molecule contains 79,32g of oxygen.
nitrogen
11,6% of 120g
[tex]m=120g*0,0,116\\m=13,92g[/tex]
One molecule contains 13,92g of nitrogen.
4 amount of substance
carbon
[tex]n=\frac{23,76g}{12,01\frac{g}{mol} }\\n=1,98mol[/tex]
The compound contains about 2 moles of carbon.
hydrogen
[tex]n=\frac{3g}{1\frac{g}{mol} }\\n=3mol[/tex]
The compound contains about 3 moles of hydrogen.
oxygen
[tex]n=\frac{79,32g}{15,999\frac{g}{mol} }\\n=4,96mol[/tex]
The compound contains about 5 moles of oxygen.
nitrogen
[tex]n=\frac{13,92g}{14,007\frac{g}{mol} }\\n=0,99mol[/tex]
The compound contains about 1 moles of nitrogen.
5. molecular formula
The formula results from the ratio of the amounts of substance.
[tex]n_{C} :n_{H} :n_{O} :n_{N} =2:3:5:1\\C_{2}H_{3}NO_{5}[/tex]
The molecular formula of the given compound is C₂H₃NO₅, and percent composition of oxygen in it is 66.1%.
How do we calculate mass from % composition?
Mass of any composition of any compound will be calculated by using the below formula as:
Mass of component = (% composition)×(mass of compound) / 100
Given mass of compound = 120g/mol
Total composition of compound (100%) = Percent composition of all components
% composition of oxygen = 100 - (19.8 + 2.50 + 11.6) = 66.1%
Moles will be calculated as:
n = W/M, where
W = given mass
M = molar mass
- For carbon atom:
Mass of Carbon component = (0.198)(120g) = 23.76g
Moles of Carbon atom = 23.76g / 12.01g/mol = 1.98mol = 2 moles
- For nitrogen atom:
Mass of Nitrogen component = (0.116)(120g) = 13.92g
Moles of Nitrogen atom = 13.92g / 14.007g/mol = 0.99mol = 1 moles
- For oxygen atom:
Mass of Oxygen component = (0.661)(120g) = 79.32g
Moles of Oxygen atom = 79.32g / 15.99g/mol = 4.96mol = 5 moles
- For hydrogen atom:
Mass of Hydrogen component = (0.025)(120g) = 3g
Moles of Hydrogen atom = 3g / 1g/mol = 3 moles
So, the molecular formula of the compound on the basis of moles of given entities is C₂H₃NO₅.
Hence required molecular formula is C₂H₃NO₅.
To know more about emperical formula, visit the below link:
https://brainly.com/question/1603500
