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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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4 years ago
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3 years ago
Determine the acid dissociation constant for a 0.020 m formic acid solution that has a ph of 2.74. formic acid is a weak monopro
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Answer:

Ka = 1.82x10⁻⁴

Explanation:

To solve this problem, we need a basis. In this case, we need the overall reaction and a ICE chart. So, let's write the overall reaction first:

HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

Now that we have the overall reaction, we need to write the ICE chart. In this way we can determine what data do we have, and what do we have left to determine:

      HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

i)        0.02                                0             0

e)       0.02-x                             x              x

The Ka expression is:

Ka = [H₃O⁺] [HCOO⁻] / [HCOOH]

Replacing the given data we have:

Ka = x² / 0.02 - x

Now, the value of x can be calculated because we already have the pH of the formic acid, and with it, we can calculate the [H₃O⁺] with the following expression:

[H₃O⁺] = 10^(-pH)

Replacing we have:

[H₃O⁺] = 10^(-2.74) = 1.82x10⁻³ M

This is the value of x, so replacing in the Ka expression, we can calculate then, the value of Ka:

Ka = (1.82x10⁻³)² / (0.02 - 1.82x10⁻³)

<h2>Ka = 1.82x10⁻⁴</h2>
7 0
3 years ago
An aqueous solution of potassium carbonate combine with a solution of calcium nitrate. What are the total and net ionic equation
romanna [79]
<span><span>K_2</span>C<span>O_3</span>(aq)+Ca(N<span>O_3</span><span>)_2</span>(aq)→ ?</span>
If we break these two reactants up into their respective ions, we get...<span><span>
K^+ </span>+ C<span>O^2_3 </span>+ C<span>a^<span>2+ </span></span>+ N<span>O_−3</span></span>
If we combine the anion of one reactant with the cation of the other and vice-versa, we get...<span>
CaC<span>O_3 </span>+ KN<span>O_3</span></span>

Now we need to ask ourselves if either of these is soluble in water. Based on solubility rules, we know that all nitrates are soluble, so the potassium nitrate is. Alternatively, we know that all carbonates are insoluble except those of sodium, potassium, and ammonium; therefore, this calcium carbonate is insoluble. This is good. It means we have a driving force for the reaction! That driving force is that a precipitate will form. In such a case, a precipitation reaction will occur, and the total equation will be...<span><span>
K_2</span>C<span>O_3</span>(aq) + Ca(N<span>O_3</span><span>)_2</span>(aq) → CaC<span>O_3</span>(s) + 2KN<span>O_3</span>(aq)</span>

To determine the net ionic equation, we need to remove all ions that appear on both sides of the equation in aqueous solution -- these ions are called spectator ions, and do not actually undergo any chemical reaction. To determine the net ionic equation, let's first rewrite the equation in terms of ions...
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The species that appear in aqueous solution on both sides of the equation (spectator ions) are... <span>
2K^+,NO_3^-</span>
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CO_3^{2-}(aq) + Ca^{2+}(aq) <span>→</span> CaCO_3(s)
5 0
3 years ago
(06.04 LC)
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Answer:

type of particles

that's d

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3 years ago
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