A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element present and the number of atoms of each element present in the sample.

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01  . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01  . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

For 2.01 moles PNBr2 we have 1.21 *10^24 molecules

P =  1.21 *10^24 atoms

N = 1.21 *10^24 atoms

Br = 2.42*10^24 atoms

Explanation:

Step 1: Data given

Number of moles of PNBr2 = 2.01 moles

Step 2: Calculate moles per element

For 1 mol PNBr2 we have 1 mol P, 1 mol N and 2 moles Br

For 2.01 moles PNBr2 we have:

P = 2.01 moles

N = 2.01 moles

Br = 2* 2.01 = 4.02 moles

Step 3: Calculate atoms of each element

For 2.01 moles PNBr2 we have 6.022 *10^23 * 2.01 = 1.21 *10^24 molecules

P = 2.01 moles = 6.022 *10^23 * 2.01 = 1.21 *10^24 atoms

N = 2.01 moles = 6.022 *10^23 * 2.01 = 1.21 *10^24 atoms

Br = 4.02 moles = 6.022 *10^23 * 4.02 = 2.42*10^24 atoms

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o