Answer:
Empirical and molecular formulas are the same, C₅H₁₀O₂.
Explanation:
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In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:
Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:
And consequently the moles:
Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:
Thus, the empirical formula, taken the nearest whole number is:
Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.
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Atomic mass of F: 19.0 g/mol
Atomic mass of S: 32.1 g/mol
1.18 g F = [1.18 g / 19.0 g / mol] = 0.062 mol F
1 g S = 1 g/ 32.1 g/mol = 0.031 mol S
Divide by 0.031
0.062 mol F / 0.031 = 2 mol F
0.031 mol S / 0.031 = 1 mol S
SF2 Then X = 2
Verification:
F2 = 2*19.0 g = 38 g F
S = 32.1 g
36 gF / 32.1 g S = 1.18 g F / g S
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
