# How to determine the molecular formula

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Sections: Chemistry

Students are faced with the tasks of deriving the chemical formula of a substance while completing a chemistry program from grades 8 to 11. In addition, this type of problem is quite common in olympiad tasks, control and measuring materials of the exam (parts B and C). The complexity range of these tasks is wide enough. As experience shows, schoolchildren often have difficulties already in the first stages of the solution when removing the molar mass of a substance.

This development proposes tasks for finding the formula of a substance, based on various parameters in the conditions. In the presented problems, various methods for finding the molar mass of a substance are given. The tasks are designed in such a way that students can learn the best methods and various solutions. The most common decision-making techniques are clearly demonstrated. For students, they are offered solved problems on the principle of increasing complexity and tasks for independent solutions.

Conclusion of the chemical formula of the substance:

(solution example)

Calculation of the molar mass of a substance

- based on mass fractions (%) of element atoms

M, where n is the number of atoms

Determine the chemical formula of the compound having the composition: sodium - 27.06%, nitrogen - 16.47%, oxygen - 57.47%. Answer:NaNO3

- based on the mass fractions (%) of the atoms of the elements and the density of the compound

M (CxNu) = D (H2); M (H2)

The relative vapor density of the organic oxygen-containing compound in oxygen is 3, 125. The mass fraction of carbon is 72%, hydrogen is 12%. Derive the molecular formula of this compound. Answer:C6H12ABOUT

- by the density of the substance in a gaseous state

M (v-va) = ρ · M (gas sampling v-va)

The relative vapor density of the limit aldehyde for oxygen is 1.8125. Derive the molecular formula of the aldehyde. Answer:C3N6ABOUT

- based on the mass fractions (%) of the atoms of the elements and the mass of the compound

M is in the ratio
or
M

Hydrocarbon contains 81.82% carbon. Weight 1 liter of this hydrocarbon (ns) is 1.964 g. Find the molecular formula of the hydrocarbon.

- by weight or volume of the starting material and combustion products

M (in-va) = Vm

The relative vapor density of an oxygen-containing organic compound in helium is 25.5. When burning 15.3 g of this substance, 20.16 l were formed. With2 and 18.9 g. N2A. Derive the molecular formula of this substance.Answer:C6H14ABOUT

An example of solving problem No. 6 on the application of the Mendeleev-Klaiperon equation is given.

The mass fraction of oxygen in the monobasic amino acid is 42.67%. Set the molecular formula of the acid.

Decision:
Calculate the molar mass of acid CnN2n (N N2) COOH
w (O) =

M acid = 75 (g / mol)
Find the number of carbon atoms in the acid molecule and establish its formula M = 12 n + 2 n + 16 + 45 = 75
14 n = 14, n = 1
M (NН2CH2 COOH) = 75 g / mol

Derive the compound formula
CnN2n (N N2) COOH

The relative density of a hydrocarbon in hydrogen having the composition: w (C) = 85.7%, w (H) = 14.3%, is 21. Derive the molecular formula of the hydrocarbon.

Given:
w (C) = 85.7%
w (H) = 14.3%
D H2 (CxNu) = 21

1. We find the relative molar mass of the hydrocarbon, based on the value of its relative density: M (CxNu) =D (N2) M (N2)

M (CxNu) = 21 · 2 = 42

m (H) = 42g. / 100 %14.3% = 6 g.
Find the amount of matter of carbon and hydrogen atoms
n (C) = 36 g: 12 g / mol = 3 mol
n (H) = 6g.: 1 g / mol = 6 mol

Answer: true formula of substance C3N6.

Derive the compound formula
CxNu?

Determine the molecular formula of an alkane if it is known that its vapor is 2.5 times heavier than argon.

Given:
Alkane vapor 2.5 times heavier than argon

Decision:
By relative density, you can find the molar mass of alkane: M(C n H 2 n + 2)= 14 n + 2 = 2.5 · M (Ar) = 100 g / mol
Whence n = 7.

Derive alkane formula
C n H2 n + 2

The mass fraction of carbon in the compound is 39.97%, hydrogen 6, 73%, oxygen 53.30%. Weight 300 ml. (ns) of this compound is 2.41 g. Derive the molecular formula of this substance.

Given:
w (C) = 39.97%
w (H) = 6.73%
w (0) = 53.30%
Vn.u. (CxHyOz) = 300 ml.
m (CxHyOz) = 2.41 g.

Decision:
For calculation, select 100g. connections. Then the mass of carbon is 39.97 g, hydrogen 6.73 g, oxygen 53.30 g.
1. Determine the amount of substance:
n (C) = 39.97 g: 12 g / mol = 3.33 mol
n (H) = 6.73 g: 1.008 g / mol = 6.66 mol
n (0) = 53.3 g: 16 g / mol = 3.33 mol
Determine the smallest common multiple - 3.33.
n (C): n (H): n (0) = 1: 2: 1
The simplest compound formula is CH2ABOUT
M (CH2O) = 30 g / mol
Determine the molar mass of the compound in the ratio:
0.3 l. - 2.41 g.
22.4 L. - x
x = (22.4 · 2.41) / 0.3 = 180
Or according to the formula M =Vmm /V
K = 180: 30 = 6
We determine the molecular formula of the compound by multiplying the stoichiometric coefficients in the simplest formula by 6.

Derive the compound formula
CxNuOz-?

What is the molecular formula of a hydrocarbon having a density of 1.97 g / l, if during the combustion of 4.4 g it formed in oxygen 6.72 l. CO2 and 7.2 g of H2O.

Given:
M (CxHy) = 4.4 g.
ρ (n.o.) = 1.97 g / l
V (CO2) = 6.72 l.
m (H2O) = 7.2 g.

Decision:
1. We find the relative molar mass of the hydrocarbon, based on the value of its relative density:
M (CxHu) =Vm ρ
M (CxHy) = 22.4 l / mol · 1.97g / l = 44g / mol
2. We write in the algebraic form the equation of the gas combustion reaction, expressing the coefficients in x and y.

We make proportions:
4.4 / 44 = 6.72 / x22.4
x = 44.6, 72 / 4.422.4 = 3
y = 44 · 7.2 / 4.4 · 9 = 8
Compound Formula C3H8, M (C3H8) = 44 g / mol
Answer: the molecular formula of compound C3H8

Print formula
CxHu -?

The compound contains 62.8% S and 37.2% F. The mass of 118 ml of this compound at 70 and 98.64 KPa is 0.51 g. Derive the formula of the compound.

Given:
w (S) = 62.8%
w (F) = 37.2%
m (CxHy) = 0.51 g
V (CxHy) = 118 ml.
T = 70
P = 98.64 kPa

1. We determine the simplest compound formula:

n (S): n (F) = 62.80 / 32: 37.2 / 19 = 1.96: 1.96 = 1: 1
The simplest formula S F

1. Find the molar mass of the compound:

M= (0, 51 · 8.31 · 280) / (98.64 · 103 · 118 · 10-6) = = 101.95 g / mol.

Therefore, the formula of compound S2 F2

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