Review for Unit 4 Assessment

Write the formula for the following compounds? L.T. #5

Carbon tetrabromide

Dioxygen tetranitride

Heptasulfur pentaiodide

Triphosporus monoselenide

CBr4

O2N4

S7I5

P3Se

Name the following compounds L.T. #4

Br2O6

C3N5

S8P4

NCl

Dibromine hexaoxide

Tricarbon pentanitride

Octasulfur tetraphosphide

Nitrogen monochloride

How many moles are in 100 g of C3N5? L.T. = #10, 11

100 g (___________)

= 0.94 moles C3N5

1 mole

From periodic table: (C3N5)

C = 12.01 g (3) = 36.03 g

N= 14.01 g (5) = + 70.05 g

molar mass = 106.08 g

106.08 g

How many molecules are in 4.1 moles of Br2O6 ? L.T. = #10, 11

= 2.468 x 1024 mc

4.1 mol (____________)6.02 x 1023 mc1 mol

What is the percent composition of each element in triphosphorus pentachloride?

/ 270.16 = 34.4 % P

From periodic table:

P = 30.97 g (3) = 92.91 g

Cl= 35.45 g (5) = + 177.25 g

molar mass = 270.16 g

P3Cl 5

/ 270.16 = 65.6 % Cl

L.T. = #8

What is the bond type for each compound?

C = 2.5 Br = 2.8

Difference = 0.3

Bond type = nonpolar covalent

CBr4

Al2S3

LiF

Al = 1.5 S = 2.5

Bond type = Polar covalent

Li = 1.0 F = 4.0

Bond type = Ionic

L.T. = #3

What might be the properties for each compound?

Nonpolar = liquid or gas, low m.p., low b.p., soluble in nonpolar substances

CBr4

LiF Ionic = solid, high m.p., high b.p., soluble in ionic substances, excellent conductor

L.T. = #2

What is the molecular shape for each compound?

tetrahedralCBr4

SO2

PCl3

AlBr3

angular

triangular pyramidal

triangular planar

L.T. = #6,7

What is the percent composition the salt in the following hydrate compound; CuSO4 * 5H2O?

Cu = 63.55 g (1) = 63.55 g

S= 32.07 g (1) = 32.07 g

O = 16.00 g (4) = + 64.00 g

molar mass salt = 159.62 g/mol / 249.72 =

63.69 % salt

L.T. = #9

H2O = 18.02 (5) = + 90.10 g/ mol

molar mass hydrate = 249.72 g/mol