A compound is found to contain 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. It has a molar mass of 135 g. What is the molecular formula?

C6H12O3

Explanation:

Data obtained from the question include:

C (carbon) = 54.5%

H (hydrogen) = 9.1%

O (oxygen) = 36.4%

To obtain the molecular formula, first, let us obtain the empirical formula. This is illustrated below

C = 54.5%

H = 9.1%

O = 36.4%

Divide by their individual molar mass.

C = 54.5/12 = 4.542

H = 9.1/1 = 9.1

O = 36.4/16 = 2.275

Divide by the smallest

C = 4.542/2.275 = 2

H = 9.1/2.275 = 4

O = 2.275/2.275 = 1

The empirical formula is C2H4O

The molecular formula is simply a multiple of the empirical formula. This is illustrated below:

Molecular formula => [C2H4O]n

The molar mass of the compound as obtained from the question is 135g/mol

Now, with the above information, we can simply calculate the value of n in order to obtain the molecular formula. This is illustrated below:

[C2H4O]n = 135

[(12x2) + (4x1) + 16]n = 135

[24 + 4+ 16]n = 135

44n = 135

Divide both side by the coefficient of n i.e 44

n = 135/44

n = 3

The molecular formula

=> [C2H4O]n

=> [C2H4O]3

=> C6H12O3

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.