A compound is composed of the elements C, H and N. a) Combustion analysis of a 5

Question

A compound is composed of the elements C, H and N. a) Combustion analysis of a 5.72 gsample ofthe compound gave 10.5 g of carbon dioxide and 1.72 gof water. What is the empirical formula of the compound? b) The molar mass is estimated at 115-130 amu. What is the molecular formula? Naturally occurring copper consists of two isotopes: Cu-63 and Cu-65. Cu-63 has an exact mass of 62.930 amu and an abundance of 69.09%. What is the mass of the Cu-65 isotope to four significant figures? A hydrocarbon is 86% carbon and 14% hydrogen by mass. What is the empirical formula of the compound?

Explanation / Answer

Answer –

a)

We are given, mass of sample = 5.72 g, mass of CO2 = 10.5 g

Mass of water = 1.72 g

Now first we need to calculate mass of C, H and N

Moles of CO2 = 10.5 g / 44 g.mol-1

                       = 0.239 moles of CO2

Moles of H2O = 1.72 g / 18.015 g.mol-1

                       = 0.0955 moles

We know, 1 moles of CO2 = 1 moles of C

So moles of C = 0.239 moles

1 moles of H2O = 2 mole of H

So, moles of H = 2*0.0955 = 0.191 moles

Mass of C = 0.239 mole * 12.011 g/mol

                 = 2.87 g

Mass of H = 0.191 mole * 1.0079 g/mol

                  = 0.192 g

So mass of N = 5.72 g – 2.87 g – 0.192 g

                       = 2.66 g of N

Moles of N = 2.66 g / 14.007 g/mol

                   = 0.190 moles

Mow from the moles of C, H and N lowest moles for N, so divide each mole by 0.190

C = 0.239 /0.190 = 1.25

H = 0.191 / 0.190 = 1

N = 0.190/0.190 = 1

Now we need to make the number fraction to whole number so multiply such number to all so we will get whole number for C

C = 1.25 *4 = 5

H = 1*4 = 4

N = 1*4 = 4

So empirical formula is C5H4N4

b) We are given, molecular formula mass = 115.13 g/mole

We know,

Molecular formula = n * empirical formula

n = molecular formula mass / empirical formula mass

   = 115.13 / 120.11

= 0.95 means 1

So molecular formula and empirical formula both are same here and that is C5H4N4

Next question –

We are given percent abundance of Cu-63= 69.09 % , atomic mass of Cu-63= 62.930 amu

Percent abundance of Cu-65 = 100 – 69.09 = 30.91 %

Now we need to calculate the mass for Cu-65

Average mass of Cu = 63.546 amu

Average mass = (percent abundance of Cu-63*atomic mass of Cu-63) + (percent abundance of Cu-65*atomic mass of Cu-65) / 100

63.546 amu = (69.09*62.930) + (30.91* x) /100

                   = 4347.8 + 30.91x

30.91x = (63.546*100) - 4347.8

             = 2006.7

x = 64.92 amu

Next question-

We are given percent of C = 86 % . percent of H = 14 %

We assume 100 g of sample

So, mass of C = 86 g and mass of H = 14 g

Moles of C = 86 g / 12.011 g/mole = 7.16

Moles of H = 14 g / 1.0079 g/mole = 13.8

Now moles of C is lowest, so we need to divided each mole by 7.16

C = 7.16 / 7.16 = 1

H = 13.8 / 7.16 = 1.94 = 2

So empirical formula is CH2

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