Firstly, find the percent composition of hydrogen in the compound (ammonium nitrate).
So we have:
2 x Nitrogen atom = 28.02 amu
4 x Hydrogen atoms = 4.04 amu
3 x Oxygen atoms = 48.00 amu
Total of: 80.06 amu or g/mol
% H = 4.04 g/mol ÷ 80.06 g/mol = 0.050462153385 = 5.05 % (sig. figs.)
Then, we take the percentage of hydrogen and multiply it by the mass of the substance.
50.00 g NH4NO3 x 0.050462153385 = 2.52310766925 = 2.52 g H
So, there is 2.52 g of hydrogen in 50.00 g of NH4NO3
NH4NO3's molar mass = 80.0gMolarity = moles of solute / liters of solution0.450 M = X / 0.500 LX = .225 molesNow we must convert X to grams because it is currently in moles.(.225 moles) / 1 * ( 80.0 grams ) / 1 mole =18 grams of NH4NO3
The percent nitrogen in NH4NO3 is 35.04%. This can be calculated by dividing the molar mass of nitrogen in the compound by the molar mass of the entire compound and then multiplying by 100.
Formula: NH4NO3 It is actually Ammonium Nitrate
There are a total of 26 atoms in one molecule of NH4NO3. This includes 2 nitrogen atoms, 4 hydrogen atoms, and 3 oxygen atoms.
To find the number of atoms of nitrogen in 20g of NH4NO3, first calculate the molar mass of NH4NO3 (80.043 g/mol). Next, divide the mass of NH4NO3 by its molar mass to find the number of moles (0.2496 mol). Finally, multiply the number of moles by Avogadro's number (6.022 x 10^23) to determine the number of atoms of nitrogen (1.50 x 10^23 atoms).
To calculate the amount of nitrogen in ammonium nitrate, we first need to determine its molecular formula, which is NH4NO3. This means that there is one nitrogen atom in each molecule. One mole of NH4NO3 weighs approximately 80 grams, and contains one mole of nitrogen which weighs about 14 grams. To find the amount of nitrogen in 35.8 kg of ammonium nitrate, we can set up a simple proportion: (35.8 kg / 80 kg) * 14 g = 6.29 g of nitrogen.
One mole of NH4 (ammonium) has one mole of nitrogen atoms and four moles of hydrogen atoms, for a total of five moles of atoms. Multiply by Avogadro's Number to convert moles of atoms to atoms.
NH4NO3's molar mass = 80.0gMolarity = moles of solute / liters of solution0.450 M = X / 0.500 LX = .225 molesNow we must convert X to grams because it is currently in moles.(.225 moles) / 1 * ( 80.0 grams ) / 1 mole =18 grams of NH4NO3
The percent nitrogen in NH4NO3 is 35.04%. This can be calculated by dividing the molar mass of nitrogen in the compound by the molar mass of the entire compound and then multiplying by 100.
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%
Formula: NH4NO3 It is actually Ammonium Nitrate
N2O is the compounds with the highest amount of nitrogen than the srno32, nh4no3 and hno3.
There are a total of 26 atoms in one molecule of NH4NO3. This includes 2 nitrogen atoms, 4 hydrogen atoms, and 3 oxygen atoms.
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.
To find the number of atoms of nitrogen in 20g of NH4NO3, first calculate the molar mass of NH4NO3 (80.043 g/mol). Next, divide the mass of NH4NO3 by its molar mass to find the number of moles (0.2496 mol). Finally, multiply the number of moles by Avogadro's number (6.022 x 10^23) to determine the number of atoms of nitrogen (1.50 x 10^23 atoms).
15 grams of nitrogen are equal to 1,071 moles.
The chemical name for NH4NO3 is ammonium nitrate. Ammonium nitrate is composed of the elements nitrogen (N), hydrogen (H) and oxygen (O). NH4NO3 has a molecular weight of 80.05-grams per mole.