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Chapter 9: Problem 148

Nicotine has the formula \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}\). Determine the mass percent of each element.

Short Answer

Expert verified
The mass percentages of each element in nicotine (\(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{N}_{2}\)) are approximately: Carbon: \(74.03\%\) Hydrogen: \(8.699\%\) Nitrogen: \(17.27\%\)

Step by step solution

01

Find the molar mass of nicotine

To determine the molar mass of nicotine, we need to add the molar mass of each element present in the compound, multiplied by the number of atoms of each element. The molar masses of carbon (C), hydrogen (H), and nitrogen (N) are approximately 12.01 g/mol, 1.008 g/mol, and 14.01 g/mol respectively. The molar mass of nicotine (\(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{N}_{2}\)) can be calculated as follows: Nicotine molar mass = (10 × 12.01 g/mol) + (14 × 1.008 g/mol) + (2 × 14.01 g/mol)
02

Calculate the molar mass of each element in nicotine

Now, we will multiply the molar mass of each element by the number of atoms of that element in nicotine. Carbon: 10 × 12.01 g/mol = 120.1 g/mol Hydrogen: 14 × 1.008 g/mol = 14.112 g/mol Nitrogen: 2 × 14.01 g/mol = 28.02 g/mol
03

Calculate mass percentages of each element

Lastly, we will divide the molar mass of each element in nicotine by the molar mass of nicotine and multiply the result by 100 to obtain the mass percentages of each element. Carbon (%): \(\frac{120.1}{162.232} × 100\% = 74.03\%\) Hydrogen (%): \(\frac{14.112}{162.232} × 100\% = 8.699\%\) Nitrogen (%): \(\frac{28.02}{162.232} × 100\% = 17.27\%\) Therefore, the mass percentages of each element in nicotine are approximately: Carbon: 74.03% Hydrogen: 8.699% Nitrogen: 17.27%

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Most popular questions from this chapter

For the reaction \(\mathrm{NH}_{3}+\mathrm{NO} \rightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}:\) (a) Balance the equation. (b) If you react \(15.0 \mathrm{~g} \mathrm{NH}_{3}\) with \(22.0 \mathrm{~g} \mathrm{NO}\) and you produce \(13.3 \mathrm{~g} \mathrm{~N}_{2}\), what is your percent yield?

Consider the unbalanced chemical equation \(\mathrm{CaC}_{2}+\mathrm{CO} \rightarrow \mathrm{C}+\mathrm{CaCO}_{3}\) When the reaction is complete, \(135.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) produced and \(38.5 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) is left over. Assuming the reaction had a \(100 \%\) yield, what were the mass of the two reactants at the beginning of the reaction

Iron(III) oxide reacts with carbon monoxide to give iron metal and carbon dioxide. If you begin the reaction with \(24.0 \mathrm{~g}\) of iron(III) oxide and \(34.0 \mathrm{~g}\) of carbon monoxide, what is the theoretical yield in grams of carbon dioxide?

Calcium reacts with nitrogen gas to form calcium nitride. If \(33.8 \mathrm{~g}\) of calcium react with \(20.4 \mathrm{~g}\) of nitrogen gas, (a) Which reactant is the limiting reagent? (b) If the reaction has a \(72.4 \%\) yield, how many grams of calcium nitride are formed?

Consider the following balanced chemical equation: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (a) How many grams of water are formed from \(5.00 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and an excess amount of \(\mathrm{O}_{2}\) ? (b) How many grams of \(\mathrm{O}_{2}\) do you need to produce \(5.00 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) ? (c) Given \(100.0 \mathrm{~g}\) of \(\mathrm{H}_{2}\), how many grams of \(\mathrm{O}_{2}\) are required to run the reaction in a stoichiometric fashion? (d) What is the theoretical yield in grams of water upon combining \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) with an excess amount of \(\mathrm{H}_{2}\) ? (e) Express the answer to part (d) in terms of the number of water molecules.
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