Chemistry, 21.06.2019 15:30 hannah5143

Determine the empirical formula of a compound containing 40.6 grams of carbon, 5.1 grams of hydrogen, and 54.2 grams of oxygen. in an experiment, the molar mass of the compound was determined to be 118.084 g/mol. what is the molecular formula of the compound? for both questions, show your work or explain how you determined the formulas by giving specific values used in calculations.

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Answer from: brainBoy480

- Empirical formula is C_2H_3O_2 and molecular formula is C_4H_6O_4 .

Solution:- An empirical formula is the simplest whole number ratio of moles of each kind of element present in it.

Grams of each element are converted to moles on dividing by atomic masses.

$C=40.6g(\frac{1mol}{12.01g})$

= 3.38 mol C

$H=5.1g(\frac{1mol}{1.01g})$

= 5.05 mol H

$O=54.2g(\frac{1mol}{16.0g})$

= 3.39 mol O

The next step is to calculate the mol ratio and for this we divide the moles of each by least one of them.

C = $\frac{3.38}{3.38}$ = 1

H = $\frac{5.05}{3.38}$ = 1.5

O = $\frac{3.39}{3.38}$ = 1

1.5 is not a whole number so we multiply all of these by 2 to get the simplest whole number ratio.

so, C = 2

H = 3

O = 2

The empirical formula is C_2H_3O_2 .

To find out the molecular formula we need to know how many empirical formula units are there in the molecular formula and for this we divide the molar mass by the empirical formula mass.

empirical formula mass = 2(12.01)+3(1.01)+2(16.00)

= 24.02+3.03+32.00

= 59.05

So, number of empirical formula units = $\frac{118.084}{59.05}$

= 2

Hence, the molecular formula of the compound will be two times of its empirical formula that is C_4H_6O_4 .

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

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Determine the empirical formula of a compound containing 40.6 grams of carbon, 5.1 grams of hydrogen...

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