Question

A compound contains only C, H, and N. Combustion of 70 mg of the compound produces 132 mg CO2 and 54 mg H2O. What is empirical formula of compound?

Answer 1

Answer:

We can write an equation for the (complete) combustion of the substance as follows:

CxHyNz+x+y2+z2O2→xCO2+y2H2O+zNO2

That might look complicated, but all we're saying is that each mole of C in the reactant will create 1 mole of CO2 as a product, and require 12 mole of O2 to do so.

The same kind of reasoning applies for H to H2O and N to NO2.

(There are several oxides of nitrogen such as N2O, but this one corresponds to complete combustion.)

We know the molar mass of CO2 is 44 gmol−1 and the molar mass of H2O is 18 gmol−1.

Find the number of moles of CO2 in the products:

n=mM=33.5×10−344=7.6×10−4 mol

(note that we had mg of product and grams for molar mass)

This is simply the value of x, the number of moles of C in the reactant.

Find the number of moles of CO2 in the products:

n=mM=41.1×10−318=2.3×10−3 mol

Now each mole of H in the reactant produces only 12 mole of H2O, so we need to double this number to find y, the number of moles of H in the reactant: y=4.6×10−3 mol

We can now find the mass of C and the mass of H in the reactant:

For C:

m=nM=7.6×10−4⋅12=9.12×10−3 g = 9.12 mg

For H:

m=nM=4.6×10−3⋅1=4.6×10−3 g = 4.6 mg

Now, if we add these we have 13.72 mg, but we know the sample was 35.0 mg, so the remainder must be N: 21.28 mg.

Find the number of moles of N:

n=mM=21.28×10−314=1.52×10−3 mol

Now we have:

x=7.6×10−4=0.76×10−3 mol

y=4.6×10−3 mol

z=1.52−3 mol

These are not neat and tidy numbers, but they look close to 34,412and112. To get them to nice clean whole numbers, multiplying by 4000 seems to make sense:

x≈3,y≈18,z≈6

Hmm, close, but we can make it simpler by dividing by 3:

x≈1,y≈6,z≈2

So our empirical formula, in its simplest form, is:

CH6N2

Explanation: mark me as brainliest

Answer 2

C3 H6 N2 .Answer:

Explanation:GIVEN equation of reaction can be written as

CxHyNz +O2 =xCO2+y/2H2O+zN

mass of co2 produce 132mg [given]

then mole of co2 produce is 0.003

then mole of c is 0.003

similarlly mole of h20 is 0.003

then,

mole of hydrogen 0.006

then simplest ratio of carbon and hydrogen is 3,6

mass of compound 70

now C3H6Nz

36+6+14z=70

14z=28

z=2

hence empirical formula of compound is C3H6N2