0.103g sample of NH4NO3 required 12.8ml of 0.101M NaOH for neutralization.what is the %purity of a sample?
NH4NO3 + NaOH ,p==> NH4OH + NaNO3 ==> NH3 + H2O + NaNO3
millimoles NaOH = M x mL = 0.101 M x 12.8 mL = 1.29 millimoles = 0.00129 moles.
For every 1 mole NaOH you had 1 mole NH4NO3; therefore, moles NH4NO3 = 0.00129.
grams NH4NO3 = moles x molar mass = 0.00128 x 80 = 0.103 g
% purity = (grams NH4NO3/g sample)*100 = ?
Well, let's crunch some numbers and make chemistry fun!
First, let's calculate the number of moles of NaOH used in the reaction. We know the volume (12.8 mL) and concentration (0.101 M), so we can use the formula:
moles = concentration x volume
moles of NaOH = 0.101 M x 0.0128 L = 0.0012952 moles
Since the balanced equation between NH4NO3 and NaOH is 1:1, we know that the number of moles of NH4NO3 is also 0.0012952 moles.
Now, let's calculate the molar mass of NH4NO3:
M(NH4NO3) = (1 x M(N)) + (4 x M(H)) + M(O) + 3 x (1 x M(H) + 1 x M(N))
M(NH4NO3) = (1 x 14.01) + (4 x 1.01) + 16.00 + 3 x (1 x 1.01 + 1 x 14.01)
M(NH4NO3) = 28.01 + 4.04 + 16.00 + 3 x (1.01 + 14.01)
M(NH4NO3) = 80.06 g/mol
Finally, let's calculate the percentage purity:
percentage purity = (mass of pure NH4NO3 / mass of sample) x 100
mass of pure NH4NO3 = moles of NH4NO3 x molar mass of NH4NO3
mass of pure NH4NO3 = 0.0012952 moles x 80.06 g/mol = 0.1037 g
percentage purity = (0.1037 g / 0.103 g) x 100 ≈ 100.7%
So, the percentage purity of the NH4NO3 sample is approximately 100.7%.
Remember, chemistry can be a blast!
To find the % purity of a sample of NH4NO3, we need to determine the amount of NH4NO3 in the sample and compare it to the total weight of the sample.
Step 1: Calculate the amount of NH4NO3 in moles.
Given:
Sample weight = 0.103 g
Volume of NaOH used for neutralization = 12.8 mL
Molarity of NaOH = 0.101 M
Molar mass of NH4NO3:
NH4NO3 = (14.01 * 2) + (1.01 * 4) + 14.01 + (16.00 * 3)
= 28.02 + 4.04 + 14.01 + 48.00
= 94.07 g/mol
Number of moles of NaOH used:
Moles of NaOH = Volume (in L) * Molarity
= 12.8 mL * (1 L / 1000 mL) * 0.101 mol/L
= 0.0012976 mol
Using the balanced equation between NH4NO3 and NaOH:
NH4NO3 + NaOH → NaNO3 + H2O + NH3
The stoichiometry of NH4NO3 and NaOH is 1:1.
So, the number of moles of NH4NO3 is also 0.0012976 mol.
Step 2: Calculate the % purity.
% purity = (moles of NH4NO3 / total weight of the sample) * 100
Total weight of the sample is given as 0.103 g.
% purity = (0.0012976 mol / 0.103 g) * 100
≈ 1.26%
Therefore, the % purity of the sample of NH4NO3 is approximately 1.26%.
To find the percent purity of a sample, we need to compare the amount of the pure substance (in this case NH4NO3) with the total amount of substance present in the sample.
Here's the step-by-step explanation:
1. First, calculate the number of moles of NaOH used in the neutralization reaction.
Moles of NaOH = volume (in L) × concentration (in mol/L)
Moles of NaOH = 12.8 ml × 0.101 mol/L = 1.2968 × 10^-3 mol
2. Next, we need to determine the stoichiometry (molar ratio) between NH4NO3 and NaOH from the balanced chemical equation. The equation is:
NH4NO3 + NaOH ⟶ NaNO3 + H2O + NH3
From the balanced equation, we can see that 1 mole of NH4NO3 reacts with 1 mole of NaOH.
3. Since the stoichiometry is 1:1, the number of moles of NH4NO3 used in the reaction is also 1.2968 × 10^-3 mol.
4. Now, we can calculate the molar mass of NH4NO3. The molar mass is:
Molar mass of NH4NO3 = (1 × Molar mass of N) + (4 × Molar mass of H) + (3 × Molar mass of O)
= (1 × 14.01 g/mol) + (4 × 1.01 g/mol) + (3 × 16.00 g/mol)
= 14.01 g/mol + 4.04 g/mol + 48.00 g/mol
= 66.05 g/mol
5. Finally, calculate the percent purity of the sample:
Percent purity = (moles of pure NH4NO3 / total moles of sample) × 100
Percent purity = (1.2968 × 10^-3 mol / (0.103 g / 66.05 g/mol)) × 100
Percent purity = (1.2968 × 10^-3 mol / 1.5661 × 10^-3 mol) × 100
Percent purity = 82.80%
Therefore, the percent purity of the sample is approximately 82.80%.