Answer:
For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Explanation:
The computation of the mass of each element is given below:
As we know that
A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol
Now we multiply each of above by the molar masses
For N
= 14.0 g/mol × 2
= 28.0 gN
For H
= 1.0 g/mol × 4
= 4.0 gN
ANd, for O
= 16.0 g/mol × 3
= 48.0 gN
Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Answer:
<h2> 162g/mol</h2>Explanation:
The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:
<em>Nicotine has the formula </em><em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>
<em>Assume that all the atoms in nicotine are present as products </em>
<h2>Solution</h2>To find the empirical formula you need to find the moles of C, H, and N in each of the compound.
Thus, the ratio of moles is:
Divide all by the smallest number: 0.20
Hence, the empirical formula is C₅H₇N
Find the mass of 1 mole of units of the empirical formula:
Total mass = 60g + 7g + 14g = 81g
Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.
Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.