Anonymous

An oxide of nitrogen contains 30.45 mass % of N.

(a)What is the empirical formula of the oxide?

(b)If the molar mass is 90 ± 5 g/mol, what is the molecular formula?

Relevance

The compound consists of N and O only

30.45% is N, therefore

69.55% is O

Divide each % by atomic mass of element

30.45/14.007 = 2.174

69.55/15.999= 4.347

Divide by smaller number

N = 1

O = 4.347/2.174 = 2

Empirical formula = NO2 Answer to Q(a)

Molar mass NO2 = 14.007+2*15.999 = 46.005g/mol

Molar mass compound = 90+/-5

(90+/-5 ) ÷ 46.005 = 2

Molecular formula = (NO2)2 = N2O4 Answer to Q(b)

Since we have a mass percent, lets say that we have 100 g of the oxide. So we now have

69.55 g of O

30.45 g of N

To find mols of each divide by the molar mass

69.55/16 = 4.3468 mol O

30.45/14 = 2.175 mol N

Add the two together and we have 6.5218 mol of the oxide. Now find our mol ratios

4.3468 / 6.5218 = 0.666 mol% O

2.175 / 6.5218 = 0.333 mol% N

So we can see that we have 2 oxygen for every 1 nitrogen so our empirical formula is

NO2

The molar mass of our empirical formula is 46 g/mol if we multiply this by 2 we get 92 g/mol witch is within the error for the molar mass of the molecular formula. So to get our molecular formula we multiply our empirical by two. So

N2O4

• Anonymous

a) You are told that the empirical formula will only have N and O

To find the empirical formula:

1. Convert percent to mass ( pretty much just replace the % sign with the mass units, in other words consider a sample of 100g of this substance)

With a mass percent of 30.45%, converting this to mass gives 30.45g for N.

Since O is the only other element the mass percent of O is 100%-30.45%=69.55%

Converting percent to mass for O gives 69.55g

2. Convert mass to moles

moles of N= 30.45g X (inverse of the molar mass of N)

= 30.45g X mol/14g=2.175 moles

moles of O=69.55g X (inverse of the molar mass of O)= 69.55g X mol/16g=4.3468 moles

3. Divide by the smallest number of moles

Right now you have found that in 100g of this substance you would have 2.175 moles of N and 4.3468 moles of O.

Since it is not convenient to write an empirical formula that looks like

N2.175 O4.3468, you divide both 2.175 and 4.3468 by 2.175 because 2.175 is the smallest number of the two mole numbers.

After dividing your empirical formula goes from N2.175 O4.3468 to

NO1.999 where the 1.999 is just the result of 4.3468/2.175.

4. Multiply your amounts of moles until there is a formula with whole numbers of atoms or just round.

In our case you can just round 1.999 to 2.

The empirical formula then becomes NO2.

b) The molar mass of NO2 is 14 + 32=46g/mol

The molecular formula has the same ratio of elements as the empirical formula. 46 times 2 is in the range of 90g/mol so multiply the empirical formula by 2 and you get the molecular formula which is N2O4.

Source(s): www.pornhub.com