Respuesta :
the complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.
Q1)
methyl butyrate (component of apple taste andsmell): C -58.80 % H- 9.87 %
O -31.33.%Express your answer as a chemical formula.
Q2)
vanillin (responsible for the taste and smellof vanilla): C - 63.15% H- 5.30 %
O - 31.55%Express your answer as a chemical formula.
Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound
C H O
mass 58.80 g 9.87 g 31.33
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 58.80/12 9.87/1 31.33/16
= 4.9 =9.87 = 1.95
then divide number of moles by least number of moles - 1.95 in this case
4.9/1.95 = 2.51 9.87/1.95 = 5.06 1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
2.51x2 = 5.02 5.06x2 = 10.12 1x2 = 2
when rounded off to the nearest whole number
C - 5
H - 10
O - 2
therefore empirical formula is C₅H₁₀O₂
Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound
C H O
mass 63.15 g 5.30 g 31.55 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.15/12 5.30/1 31.55/16
=5.26 =5.30 =1.97
divide the number of moles by the least number of moles - 1.97
5.26/1.97 5.30/1.97 1.97/1.97
=2.67 = 2.69 = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
2.67x3 = 8.01 2.69x3 = 8.07 1x3 = 3
rounded off to the nearest whole numbers
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.
Q1)
methyl butyrate (component of apple taste andsmell): C -58.80 % H- 9.87 %
O -31.33.%Express your answer as a chemical formula.
Q2)
vanillin (responsible for the taste and smellof vanilla): C - 63.15% H- 5.30 %
O - 31.55%Express your answer as a chemical formula.
Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound
C H O
mass 58.80 g 9.87 g 31.33
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 58.80/12 9.87/1 31.33/16
= 4.9 =9.87 = 1.95
then divide number of moles by least number of moles - 1.95 in this case
4.9/1.95 = 2.51 9.87/1.95 = 5.06 1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
2.51x2 = 5.02 5.06x2 = 10.12 1x2 = 2
when rounded off to the nearest whole number
C - 5
H - 10
O - 2
therefore empirical formula is C₅H₁₀O₂
Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound
C H O
mass 63.15 g 5.30 g 31.55 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.15/12 5.30/1 31.55/16
=5.26 =5.30 =1.97
divide the number of moles by the least number of moles - 1.97
5.26/1.97 5.30/1.97 1.97/1.97
=2.67 = 2.69 = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
2.67x3 = 8.01 2.69x3 = 8.07 1x3 = 3
rounded off to the nearest whole numbers
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
