Respuesta :
This is a tricky question. All that matters are ratios of percentages, not percentages themselves. So no one should directly compare 27.2 with 42.9. We must and shall compare the ratios (27.2 to 72.8) and (42.9 to 57.1).
Take them both down to 1 to and see what happens.
Working out the formulas knowing atomic masses is a bit beside the point; this is how people first DISCOVERED the idea of atomic mass.
A
Carbon Oxygen
27.2g 72.8g (100-27.2)
Moles 27.2/12 72.8/16
2.27 4.55
Ratio 1 2
Do the same with the other
Explanation:
It is known that a mole of carbon atom contains a mass of 12 g and a mole of oxygen atom contains a mass of 16 g.
As it is given that first compound has 27.2 g C. So, amount of oxygen present in this compound will be as follows.
(100 - 27.2) g
= 72.8 g of O
Hence, calculate the moles of C and O as follows.
No. of moles of C = [tex]\frac{mass}{\text{molar mass of C}}[/tex]
= [tex]\frac{27.2 g}{12 g/mol}[/tex]
= 2.26 mol
No. of moles of O = [tex]\frac{mass}{\text{molar mass of O}}[/tex]
= [tex]\frac{72.8 g}{16 g/mol}[/tex]
= 4.55 mol
Therefore, mole ratios of both C and O will be calculated as follows.
C = [tex]\frac{ 2.26}{ 2.26}[/tex]
= 1
O = [tex]\frac{4.55}{ 2.26}[/tex]
= 2
Hence, empirical formula of the first compound is [tex]CO_{2}[/tex].
Since, it is also given that mass of carbin in another compound is 42.9 g. So, mass of oxygen present will be as follows.
(100 - 42.9) g
= 57.1 g
Hence, calculate the moles of C and O as follows.
No. of moles of C = [tex]\frac{mass}{\text{molar mass of C}}[/tex]
= [tex]\frac{42.9 g}{12 g/mol}[/tex]
= 3.57 mol
No. of moles of O = [tex]\frac{mass}{\text{molar mass of O}}[/tex]
= [tex]\frac{57.1 g}{16 g/mol}[/tex]
= 3.57 mol
Therefore, mole ratios of both C and O will be calculated as follows.
C = [tex]\frac{3.57}{3.57}[/tex]
= 1
O = [tex]\frac{3.57}{ 3.57}[/tex]
= 1
Hence, empirical formula of the second compound is CO.
Therefore, carbon and oxygen combine to form 2 different compounds in small whole numbers.
