Respuesta :
First we need to determine the amount of oxygen present.
%O = 100 - 20 -1 -12 = 67%
Thus
C = 20%
H =1%
N= 12%
O = 67%
since we are not given if the given percentages are in mass or mole, we will assume that it is by mass. therefore with a basis of 100 g of compound the mole percentages are
C = 0.22
H = 0.13
N = 0.11
O = 0.54
dividing everything by the lowest percentage (0.11) the empirical formula is
C2HNO5
%O = 100 - 20 -1 -12 = 67%
Thus
C = 20%
H =1%
N= 12%
O = 67%
since we are not given if the given percentages are in mass or mole, we will assume that it is by mass. therefore with a basis of 100 g of compound the mole percentages are
C = 0.22
H = 0.13
N = 0.11
O = 0.54
dividing everything by the lowest percentage (0.11) the empirical formula is
C2HNO5
Answer: C. [tex]C_2HNO_5[/tex]
Explanation:
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 = 20 g
Mass of H= 1 g
Mass of N= 12 g
Mass of oxygen = 100-(20+1+12)= 67 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{20g}{12g/mole}=1.7moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1g}{1g/mole}=1mole[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{12g}{14g/mole}=0.8moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{67g}{16g/mole}=4.2moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{1.7}{0.8}=2[/tex]
For H =[tex]\frac{1}{0.8}=1[/tex]
For N =[tex]\frac{0.8}{0.8}=1[/tex]
For O =[tex]\frac{4.2}{0.8}=5[/tex]
The ratio of C : H : N : O= 2 : 1 : 1 : 5
Hence the empirical formula is [tex]C_2HNO_5[/tex]
