How many moles (of molecules or formula units) are in each sample? A) 32.66 g CF2Cl2
B) 24.2 kg Fe (NO3) 2
C) 0.5912 g C8H18
D) 104 kg CaO
Express all answers to 4 sig figs

A = 1.627 ×10²³ molecules of CF₂Cl₂.

B = 809.4 ×10²³ molecules of Fe(NO₃)₂.

C = 0.03117 ×10²³ molecules of C₈H₁₈.

D= 1116 ×10²³ molecules of CaO.

Explanation:

A)

Given data:

Mass of CF₂Cl₂ = 32.66 g

Number of moles = ?

Number of molecules = ?

Solution:

Number of moles = mass/ molar mass

Number of moles = 32.66 g/ 120.9 g/mol

Number of moles = 0.2701 mol

Number of molecules:

1 mole = 6.022×10²³ molecules

0.2701 mol×6.022×10²³ molecules / 1mol = 1.627 ×10²³ molecules of CF₂Cl₂.

B) Given data:

Mass of Fe(NO₃)₂ = 24.2 Kg  (24.2×1000 = 24200 g)

Number of moles = ?

Number of molecules = ?

Solution:

Number of moles = mass/ molar mass

Number of moles = 24200 g/ 180 g/mol

Number of moles = 134.4 mol

Number of molecules:

1 mole = 6.022×10²³ molecules

134.4 mol×6.022×10²³ molecules / 1mol =809.4 ×10²³ molecules of Fe(NO₃)₂.

C) Given data:

Mass of C₈H₁₈= 0.5912 g

Number of moles = ?

Number of molecules = ?

Solution:

Number of moles = mass/ molar mass

Number of moles = 0.5912 g/ 114.22 g/mol

Number of moles = 0.005176 mol

Number of molecules:

1 mole = 6.022×10²³ molecules

0.005176 mol×6.022×10²³ molecules / 1mol =0.03117 ×10²³ molecules of C₈H₁₈.

D) Given data:

Mass of CaO = 104 Kg  (104×1000 = 104,000 g)

Number of moles = ?

Number of molecules = ?

Solution:

Number of moles = mass/ molar mass

Number of moles = 104,000 g/ 56.1 g/mol

Number of moles = 1854 mol

Number of molecules:

1 mole = 6.022×10²³ molecules

1854 mol×6.022×10²³ molecules / 1mol = 1116 ×10²³ molecules of CaO.

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

gas and precipitate r the correct answers