NH4NO3
N - 14.0 X 2 = 28
H4 - 1.0 X 4 = 4
O3 - 16.0 X 3 = 48
Total molecular mass = 80
4/80 = 5%
# of AtomsAt Wt.Total Wt.Nitrogen214.006728.0134Hydrogen41.007944.03176Oxygen315.999447.9982Total Molecular weight80.04% of Nitrogen=28.01/80.04 = 35%
To find the number of moles in 8.0g of ammonium nitrate, you need to divide the given mass by the molar mass of ammonium nitrate (NH4NO3). The molar mass of NH4NO3 is 80.05 g/mol. Therefore, 8.0g ÷ 80.05 g/mol = 0.1 moles of ammonium nitrate.
The molar mass of ammonium nitrate (NH4NO3) is approximately 80.04 g/mol.
To find the number of moles in 8.0g of ammonium nitrate, you need to divide the given mass by the molar mass of the compound. The molar mass of ammonium nitrate (NH4NO3) is 80.04 g/mol. Therefore, 8.0g ÷ 80.04 g/mol = 0.1 moles of ammonium nitrate.
Ammonium Nitrate = NH4NO3 or N2H4O3 the total mass of a ammonium nitrate molecule is as follows: 2*14u + 4*1u + 3*16u = 28u + 4u + 48u = 80u now, the total mass of Nitrogen in one ammonium nitrate is 2*14u = 28u. Then, we divide 28 (the mass of Nitrogen) by 80 (total mass)= 28/80 = 0.35, which is the ratio for Nitrogen mass divided by total mass. then, we get the 48.5 grams (total mass) and multiply it by this ratio (0.35): 48.5 * 0.35 = 16.975 grams of Nitrogen in 48.5 grams of Ammonium Nitrate.
# of AtomsAt Wt.Total Wt.Nitrogen214.006728.0134Hydrogen41.007944.03176Oxygen315.999447.9982Total Molecular weight80.04% of Nitrogen=28.01/80.04 = 35%
The chemical formula is (NH4)HPO4 (if you think to this phosphate). The molar mass is 132,07. The percentages of the elements are: - hydrogen: 6,86997 % - nitrogen: 21,21196 % - oxygen: 48,45843 % - phosphorous: 23,45964 %
The chemical formula of ammonium nitrate is NH4NO3.The molecular mass is 80,052 g.
To find the number of moles in 8.0g of ammonium nitrate, you need to divide the given mass by the molar mass of ammonium nitrate (NH4NO3). The molar mass of NH4NO3 is 80.05 g/mol. Therefore, 8.0g ÷ 80.05 g/mol = 0.1 moles of ammonium nitrate.
Ammonium hydrogen sulfate, (NH4)HSO4, contains one nitrogen atom in the ammonium ion. To calculate the percentage of nitrogen by mass, you would find the molar mass of nitrogen in the compound and divide it by the molar mass of the entire compound, then multiply by 100 to get the percentage.
The molar mass of ammonium nitrate (NH4NO3) is approximately 80.04 g/mol.
To find the number of moles in 8.0g of ammonium nitrate, you need to divide the given mass by the molar mass of the compound. The molar mass of ammonium nitrate (NH4NO3) is 80.04 g/mol. Therefore, 8.0g ÷ 80.04 g/mol = 0.1 moles of ammonium nitrate.
Ammonium Nitrate = NH4NO3 or N2H4O3 the total mass of a ammonium nitrate molecule is as follows: 2*14u + 4*1u + 3*16u = 28u + 4u + 48u = 80u now, the total mass of Nitrogen in one ammonium nitrate is 2*14u = 28u. Then, we divide 28 (the mass of Nitrogen) by 80 (total mass)= 28/80 = 0.35, which is the ratio for Nitrogen mass divided by total mass. then, we get the 48.5 grams (total mass) and multiply it by this ratio (0.35): 48.5 * 0.35 = 16.975 grams of Nitrogen in 48.5 grams of Ammonium Nitrate.
The chemical formula for ammonium phosphate is (NH4)3PO4. To calculate the mass percent of hydrogen, you first need to calculate the molar mass of the entire compound and then determine the contribution of hydrogen to the total mass. The molar mass of (NH4)3PO4 is 149.09 g/mol, and the total mass of hydrogen in the compound is 3*(1.01 g/mol) = 3.03 g. Therefore, the mass percent of hydrogen in ammonium phosphate is (3.03 g / 149.09 g) * 100% = 2.03%.
One mole of ammonium nitrate is equal to its molar mass, which is approximately 80.04 grams. This quantity represents Avogadro's number of individual ammonium nitrate molecules.
Mass of Nitrogen: 14g/mol Mass of Ammonium Nitrate (NH4NO3): 14 + 1x4 + 14 + 16x3 = 80g/mol ∴ % of Nitrogen in Ammonium Nitrate = 14/80 * 100 = 17.5%
The molecular formula of the compound is NH4NO3, which is ammonium nitrate. This compound contains nitrogen, hydrogen, and oxygen elements, with a molecular mass of approximately 63.008 amu.