Ammonia contains about 82.35 % nitrogen by mass.
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
Ammonia (NH3) has a percentage composition of 82.35% nitrogen and 17.65% hydrogen.
To find the mass of nitrogen needed to make 34 g of ammonia, we first need to calculate the molar mass of ammonia (NH3), which is 17 g/mol. From this, we can see that 1 mole of ammonia contains 1 mole of nitrogen. Therefore, the mass of nitrogen needed would also be 34 g.
# of AtomsAt Wt.Total Wt.Nitrogen214.006728.0134Hydrogen41.007944.03176Oxygen315.999447.9982Total Molecular weight80.04% of Nitrogen=28.01/80.04 = 35%
The molar mass of guanidine is 59.07 g/mol (C(NH2)3). Nitrogen contributes 42.04 g/mol to this molar mass. Therefore, the mass percentage of nitrogen in guanidine is approximately 71.2%.
To calculate the percentage composition by mass of ammonia (NH₃), first determine the molar mass of ammonia by adding the atomic masses of its constituents: nitrogen (N) has a mass of about 14.01 g/mol, and hydrogen (H) has a mass of about 1.01 g/mol. Since ammonia contains one nitrogen atom and three hydrogen atoms, its molar mass is approximately 17.04 g/mol. The percentage composition by mass of nitrogen in ammonia is calculated as (mass of N / molar mass of NH₃) × 100, which gives about 82.24%, while the percentage composition of hydrogen is (mass of H / molar mass of NH₃) × 100, resulting in about 17.76%.
The richest source of nitrogen on a mass percentage is the atmosphere, where it makes up about 78% of the air we breathe. Nitrogen gas (N2) is abundant in the atmosphere but not readily available for use by most organisms in that form.
it can be calculated using the formula percentage composition of N =Gram molecular mass of nitrogen in the compound/ Gram molecular mass of compound *100
In ammonia (NH3), the molar mass is 17 g/mol. To find the mass of nitrogen in 125 g of ammonia, first, calculate the number of moles of ammonia in 125 g. Then, multiply the moles of ammonia by the molar ratio of nitrogen in ammonia (1 mol of N for every 1 mol of NH3), and finally, multiply by the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen. This will give the mass of nitrogen in 125 g of ammonia.
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
Mass of ammonia is 17. Mass of Hydrogen in it is 3. So the percentage of hydrogen by mass is 17.64%
The molar mass of ammonia (NH3) is 17.03 g/mol. Hydrogen's molar mass is 1.01 g/mol. The mass of hydrogen in ammonia is 3.03 g (3*1.01). To find the mass percentage of hydrogen in ammonia: (3.03 g / 17.03 g) * 100% ≈ 17.8%.
The maximum percentage of nitrogen by mass is about 78.08%, which is the composition of nitrogen gas (N2) in the atmosphere.
To calculate the mass of ammonia formed, first write out the balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia: N₂ + 3H₂ → 2NH₃ Next, calculate the moles of nitrogen in 3.80 g using the molar mass of nitrogen (N₂). Then use the mole ratio from the balanced equation to determine the moles of ammonia formed. Finally, convert the moles of ammonia to grams using the molar mass of ammonia (NH₃) to find the mass formed.
Ammonia (NH3) has a percentage composition of 82.35% nitrogen and 17.65% hydrogen.
To find the mass of nitrogen needed to make 34 g of ammonia, we first need to calculate the molar mass of ammonia (NH3), which is 17 g/mol. From this, we can see that 1 mole of ammonia contains 1 mole of nitrogen. Therefore, the mass of nitrogen needed would also be 34 g.
The percentage of nitrogen in N2O4 is 63.6%. This is determined by dividing the molar mass of nitrogen in N2O4 by the molar mass of the compound and then multiplying by 100.