Question

An oxide of nitrogen was found by elemental analysis to contain 30.4% nitrogen and 69.6% oxygen. If 23.0 g of this gas were found to occupy 5.6 L at STP, what are the empirical and molecular formulas for this oxide of nitrogen?

Express the empirical and molecular formulas for this oxide of nitrogen, separated by a comma.

Answer 1

Step 1: find molar mass

we have:

P = 1.0 atm

V = 5.6 L

T = 273.0 K

find number of moles using:

P * V = n*R*T

1 atm * 5.6 L = n * 0.08206 atm.L/mol.K * 273 K

n = 0.2499 mol

mass of solute = 23.0 g

we have below equation to be used:

number of mol = mass / molar mass

0.2499 mol = (23.0 g)/molar mass

molar mass = 92.05 g/mol

step 2: find empirical and molecular formula

we have mass of each elements as:

N: 30.4 g

O: 69.6 g

Divide by molar mass to get number of moles of each:

N: 30.4/14.01 = 2.1699

O: 69.6/16.0 = 4.35

Divide by smallest to get simplest whole number ratio:

N: 2.1699/2.1699 = 1

O: 4.35/2.1699 = 2

So empirical formula is:NO?

Molar mass of NO2 = 1*MM(N) + 2*MM(O)

= 1*14.01 + 2*16.0

= 46.01 g/mol

Now we have:

Molar mass = 92.05 g/mol

Empirical formula mass = 46.01 g/mol

Multiplying factor = molar mass / empirical formula mass

= 92.05/46.01

= 2

So molecular formula is:N?O?

Answer: So molecular formula is:N?O?

Answer: NO?, N?O?