What is the number of moles in 432 g Ba(NO3)2? (Ba= 137.3 amu; N= 14.0 amu; O= 16.0 amu)?

Question

what is the number of moles in 432 g Ba(NO3)2? (Ba= 137.3 amu; N= 14.0 amu; O= 16.0 amu)

a. 0.237 mol

b. 0.605 mol

c. 1.65 mol

d. 3.66 mol

Answer 1

there is barium, two nitrogens, and six oxygens

We have (137.3) + (2 x 14.0) + (6 x 16.0) = 261.3 amu total or essentially 261.3 grams per MOLE of this compound.

We need to know what fraction of 432g is to 261.3g. Find moles:

432 grams x (1 mole/261.3grams) =

C. 1.65 moles to three significant figures (from the 14.0 or 16.0 grams)

Answer 2

You start by calculating the molar mass of your compound.  Start with the nitrate (NO3):  N + 3 O = 14 + 3(16) = 62.  You have two nitrates ((NO3)2), so 62 * 2 = 124.  Add your Barium:  124 + 137.3 = 261.3.  Units on this number are grams per mole, as we used the molar mass of each individual component to calculate the total molar mass.  So you have 432 g of something that is 261.3 g/mol.  (432 g) / (261.3 g/mol) = 1.65 mol, c.

Answer 3

The atomic mass unit is used to measure of the mass of atoms and is defined as 1/(Avogadro's number) of a gram.

Therefore, 1 mole of:

Ba = 137.3 g

N = 14.0 g

O = 16.0 g

1 mole of Ba(NO3)2 contains:

1 moles Ba = 137.3 g

2 moles N = 14.0 * 2 = 28.0 g

6 moles O = 16.0 * 6 = 96.0 g

1 mole of Ba(NO3)2 is 261.3g

432 g of Ba(NO3)2 contains:

432 g / (261.3 g/mol) = 1.65 mol

Answer: c