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inn [45]
4 years ago
6

Analysis of a sample of a compound, containing only carbon, nitrogen, hydrogen, and oxygen, determined that it contained 20.0% C

, 6.7% H, 46.6% N and the balance O. What is the empirical formula of the compound?
Chemistry
1 answer:
choli [55]4 years ago
6 0

Answer:

CH₄N₂O

Explanation:

Let's assume 100 g of the compound, thus, the mass of each substance is it percent multiplied by 100:

C = 20.0 g (20% = 0.20, and 0.20*100 = 20.0)

H = 6.7 g

N = 46.6 g

O = 100 - (20+6.7+46.6) = 26.7 g

The number of moles of each compound is the mass divided by the molar mass. The molar masses are C = 12.011 g/mol, H = 1.00794 g/mol, N = 14.01 g/mol, and O = 15.999 g/mol, so:

nC = 20/12.011 = 1.67 mol

nH = 6.7/1.00794 = 6.65 mol

nN = 46.6/14.01 = 3.33 mol

nO = 26.7/15.999 = 1.67 mol

The empirical formula is the formula with the minimum possible number of the moles of each compound, which must be proportional to the percent of each one. So, we must divide each number of moles for the smallest, 1.67:

C = 1.67/1.67 = 1

H = 6.65/1.67 = 4

N = 3.33/1.67 = 2

O = 1.67/1.67 = 1

So, the empirical formula is CH₄N₂O.

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Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

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Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

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If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
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Explanation:

Step 1: Write the balanced equation

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Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

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2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

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