A compound consists of 65.45% C, 5.492% H, and 29.06% O on a mass basis and has a molar mass of 110 g/mol. Determine the molecular formula of the compound.

The molecular formula is C6H6O2

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100 grams

The compound has:

65.45 % C = 65.45 grams

5.492 % H = 5.49 grams

29.06 % O = 29.06 grams

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.00 g/mol

Molar mass = 110 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 65.45 grams / 12.01 g/mol

Moles C = 5.450 moles

Moles H = 5.49 grams / 1.01 g/mol

Moles H = 5.44 moles

Moles O = 29.06 grams / 16.00 g/mol

Moles O = 1.816 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

Moles C = 5.450 moles / 1.816 = 3

Moles H = 5.44 moles / 1.816 = 3

Moles O = 1.816/1.816 = 1

The empirical formula is C3H3O

The molar mass of this formula is 55 g/mol

We have to multiply the empirical formula by n

n = 110/55 = 2

2*(C3H3O) = C6H6O2

The molecular formula is C6H6O2

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

d. temperate deciduous forest