Answer:
2.5 moles of NaCl
Explanation:
The balanced chemical reaction equation is shown in the image. Since it takes 2 moles of Hydrochloric acid to form two moles of sodium. Chloride, then 2.5 moles of hydrochloric acid should also form 2.5 moles of sodium chloride according to the balanced reaction equation.
This an incomplete question, here is a complete question.
Calculate the empirical formula for each stimulant based on its elemental mass percent composition.
a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%
b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%
Answer:
(a) The empirical formula for the given compound is
(b) The empirical formula for the given compound is
Explanation:
<u>Part A: nicotine </u>
We are given:
Percentage of C = 74.03 %
Percentage of H = 8.70 %
Percentage of N = 17.27 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 74.03 g
Mass of H = 8.70 g
Mass of N = 17.27 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Nitrogen =
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 1.23 moles.
For Carbon =
For Hydrogen =
For Nitrogen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 5 : 7 : 1
The empirical formula for the given compound is
<u>Part B: caffeine</u>
We are given:
Percentage of C = 49.48 %
Percentage of H = 5.19 %
Percentage of N = 28.85 %
Percentage of O = 16.48 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 49.48 g
Mass of H = 5.19 g
Mass of N = 28.85 g
Mass of O = 16.48 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Nitrogen =
Moles of Oxygen =
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 1.03 moles.
For Carbon =
For Hydrogen =
For Nitrogen =
For Nitrogen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N : O = 4 : 5 : 2 : 1
The empirical formula for the given compound is