The empirical formula : CH₃
<h3>Further explanation</h3>
Given
2.5 g sample
2.002 g Carbon
Required
The empirical formula
Solution
Mass of Hydrogen :
= 2.5 - 2.002
= 0.498
Mol ratio C : H :
C : 2.002/12 = 0.167
H : 0.498/1 = 0.498
Divide by 0.167 :
C : H = 1 : 3
Answer:
C₆H₈O₆
Explanation:
First off, the<u> percent of oxygen by mass</u> of vitamin C is:
- 100 - (40.9+4.58) = 54.52 %
<em>Assume we have one mol of vitamin C</em>. Then we would have <em>180 grams</em>, of which:
- 180 * 40.9/100 = 73.62 grams are of Carbon
- 180 * 4.58/100 = 8.224 grams are of Hydrogen
- 180 * 54.52/100 = 98.136 grams are of Oxygen
Now we <u>convert each of those masses to moles</u>, using the <em>elements' respective atomic mass</em>:
- C ⇒ 73.62 g ÷ 12 g/mol = 6.135 mol C ≅ 6 mol C
- H ⇒ 8.224 g ÷ 1 g/mol = 8.224 mol H ≅ 8 mol H
- O ⇒ 98.136 g ÷ 16 g/mol = 6.134 mol O ≅ 6 mol O
So the molecular formula for vitamin C is C₆H₈O₆
One molecule of water contains two atoms of hydrogen and one atom of oxygen, the atomicity of water is three.
Density=mass/ volume so you solve for volume and get 461.96 mL
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
