Work out the following empirical formulas using the values from the ta

1.	Work out the following empirical formulas using the values from the table(a)The smell of a pineapple is caused by ethyl butanoate. A sample is known to containonly 0.180 g of carbon, 0.030 g of hydrogen and 0.080 g of oxygen. What is theempirical formula of ethyl butanoate? (b)A 300 g sample is known to contain only carbon, hydrogen and oxygen. Thepercentage of carbon is found to be exactly the same as the percentage of oxygen.The percentage of hydrogen is known to be 5.99%. What is the empirical formula ofthe compound?c)All compositions are by mass. 29.4% Vanadium, 9.2% Oxygen, 61.4% Chlorine.
Answer» Answer: a) C3H6O B) C4H6O3 c) VOCl3 Explanation: a) Carbon. Hydrogen Oxygen 0.180/12.0 0.030/1.0 0.080/16.0 = 0.015 =0.030 =0.005 = 0.015/0.005 =0.030/0.005 =0.005/0.005 = 3 =6 = 1 Therefore, the empirical formula is C3H6O b) Hydrogen= 17.97g Oxygen= Carbon= (300-17.97)/2= 141.01g Carbon. Hydrogen Oxygen 141.0/12.0 17.97/1.0 141.0/16.0 = 11.75 =17.97 =8.81 =11.75/8.81 =17.97/8.81 =8.81/8.81 = 1.33 X3 =2.03 X 3 = 1 = 4. =6. =3 Therefore, the empirical formula is C4H6O3 C) Hydrogen= 17.97g Oxygen= Carbon= (300-17.97)/2= 141.01g VANADIUM Oxygen Chlorine 29.4/50.9 9.2/16.0 61.4/35.5 = 0.58 =0.58 =1.73 =0.58/0.58 =0.58/0.58 =1.73/0.58 = 1 =1 = 2.98 -> 3 Therefore, the empirical formula is VOCl3

Work out the following empirical formulas using the values from the table(a)The smell of a pineapple is caused by ethyl butanoate. A sample is known to containonly 0.180 g of carbon, 0.030 g of hydrogen and 0.080 g of oxygen. What is theempirical formula of ethyl butanoate? (b)A 300 g sample is known to contain only carbon, hydrogen and oxygen. Thepercentage of carbon is found to be exactly the same as the percentage of oxygen.The percentage of hydrogen is known to be 5.99%. What is the empirical formula ofthe compound?c)All compositions are by mass. 29.4% Vanadium, 9.2% Oxygen, 61.4% Chlorine.

Answer»

Answer:

a) C3H6O B) C4H6O3 c) VOCl3

Explanation:

a) Carbon. Hydrogen Oxygen

0.180/12.0 0.030/1.0 0.080/16.0

= 0.015 =0.030 =0.005

= 0.015/0.005 =0.030/0.005 =0.005/0.005

= 3 =6 = 1

Therefore, the empirical formula is C3H6O

b) Hydrogen= 17.97g

Oxygen= Carbon= (300-17.97)/2= 141.01g

Carbon. Hydrogen Oxygen

141.0/12.0 17.97/1.0 141.0/16.0

= 11.75 =17.97 =8.81

=11.75/8.81 =17.97/8.81 =8.81/8.81

= 1.33 X3 =2.03 X 3 = 1

= 4. =6. =3

Therefore, the empirical formula is C4H6O3

C) Hydrogen= 17.97g

Oxygen= Carbon= (300-17.97)/2= 141.01g

VANADIUM Oxygen Chlorine

29.4/50.9 9.2/16.0 61.4/35.5

= 0.58 =0.58 =1.73

=0.58/0.58 =0.58/0.58 =1.73/0.58

= 1 =1 = 2.98 -> 3

Therefore, the empirical formula is VOCl3