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# stoichiometry

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How many molecules of NH 4 Br are in 3.921 g of NH 4 Br ?

Mar 9, 2020

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Convert from grams into molecules:
3.921 g of NH_4Br (ammonium bromide)

Calculate the molecular mass of NH_4Br:
97.943 u

Lookup the molecular mass, m, formula:
m = sum _iN_im_i
Use the chemical formula to count the number of atoms, N_i, for each element:
| number
of atoms
Br (bromine) | 1
H (hydrogen) | 4
N (nitrogen) | 1

Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table:
| number
of atoms | atomic
mass/u
Br (bromine) | 1 | 79.904
H (hydrogen) | 4 | 1.008
N (nitrogen) | 1 | 14.007
Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m:

| | number
of atoms | atomic
mass/u | mass/u
Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904
H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032
N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007

m = 79.904 u + 4.032 u + 14.007 u = 97.943 u |

Since 1 u (unified atomic mass unit) corresponds to 1 g/mol (gram per mole), the molar mass of NH_4Br equals 97.943 g/mol. The number of moles equals the mass divided by the molar mass of NH_4Br:
(3.921 g)/1 × (1 mol)/(97.943 g) = 0.040033 mol

Look up Avogadro's constant, N_A, to find the number of molecules in a mole:
N_A = 6.022×10^23 molecules/mol

Multiply the number of moles by N_A to convert to molecules:

(0.040033 mol)/1 × (6.022×10^23 molecules)/(1 mol) = 2.411×10^22 molecules

Mar 9, 2020
edited by Guest  Mar 9, 2020