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IceJOKER [234]
3 years ago
13

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

f CO2 and 6.125 mg of H2O. What is the empirical formula for nicotine?
Chemistry
1 answer:
Gnom [1K]3 years ago
7 0

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

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Answer:

The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.

Explanation:

You know  the balanced reaction:

4 NA + O₂ ⟶ 2 Na₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:

  • Na: 4 moles
  • O₂: 1 mole
  • Na₂O: 2 moles

Being:

  • Na: 23 g/ole
  • O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

  • Na: 23 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • Na₂O: 2*23 g/mole +16 g/mole=  62 g/mole

Then by stoichiometry of the reaction they react and are produced:

  • Na: 4 moles* 23 g/mole= 92 g
  • O₂: 1 mole*32  g/mole= 32 g
  • Na₂O: 2 moles* 62 g/mole= 124 g

Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?

massofNa_{2}O =\frac{4 grams of Na*124 gramsofNa_{2}O }{92 grams of Na}

mass of Na₂O=5.39 g

<em><u>The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.</u></em>

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