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Maru [420]
4 years ago
15

Anabolic steroids are sometimes used illegally by athletes to increase muscle strength. A forensic chemist analyzes some tablets

suspected of being a popular steroid.
He determines the substance contains only C,H and O and has a molar mass of 300.42g/mol.

When a 1.200g sample is subjected to combustion analysis, 3.516g of CO2 and 1.007g H2O are collected.a) What is the empirical formula of the unknown substance in the tablets?b)What is the molecular formula of the unknown substance in the tablets?
Chemistry
1 answer:
11111nata11111 [884]4 years ago
6 0

Answer:

The empirical formula is = C_{10}H_{14}O

The formula of steroid = C_{20}H_{28}O_2

Explanation:

Mass of water obtained = 1.007 g

Molar mass of water = 18 g/mol

Moles of H_2O = 1.007 g /18 g/mol = 0.0559 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.0559 = 0.1119 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.1119 x 1.008 = 0.1128 g

Mass of carbon dioxide obtained = 3.516 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of CO_2 = 3.516 g  /44.01 g/mol = 0.0799 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.0799 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.0799 x 12.0107 = 0.9595 g

Given that the steroids only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 1.200 g

Mass of O in sample = 1.200 - 0.9595 - 0.1128 = 0.1277 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.1277 / 15.999  = 0.00798 moles

Taking the simplest ratio for H, O and C as:

0.1119 : 0.00798 : 0.0799

= 14 : 1 : 10

The empirical formula is = C_{10}H_{14}O

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 10×12 + 14×1 + 16= 150 g/mol

Molar mass = 300.42 g/mol

So,  

Molecular mass = n × Empirical mass

300.42 = n × 150

⇒ n = 2

The formula of steroid = C_{20}H_{28}O_2

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Answer:

1. Theoretical yield of NaOH is 22.72 g

2. Percentage yield of NaOH = 22.14%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaHCO₃ —> NaOH + CO₂

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.

Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.

Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e

Xmol of NaHCO₃ = 25 / 44.01

Xmol of NaHCO₃ = 0.568 mole

1. Determination of the theoretical yield of NaOH.

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.

Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.

Thus, the theoretical yield of NaOH is 22.72 g

2. Determination of the percentage yield of NaOH.

Theoretical yield of NaOH = 22.72 g

Actual yield of NaOH = 5.03 g

Percentage yield of NaOH =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.03 / 22.72 × 100

Percentage yield of NaOH = 22.14%

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