I can't understand how to solve it or what conversion factor to use 🙁 please help im studying for a test!!
You would use Avagadro’s number (6.02x10^23) to solve this. Here’s how:
8.50x10^22 molecules of NH3 x 1 mole of NH3/6.02x10^23 molecules of NH3 = 0.14 moles
Then you convert from moles to grams. First we need the molecular mass of NH3. Nitrogen’s weight is about 14 and hydrogens is about 1, but we have three hydrogens so we multiply 1x3=3. We add 14 and 3 to give us 17 grams per mole for NH3. Now we take this number to do our conversion:
0.14 moles of NH3 x 17 grams/mole = 2.38 grams of NH3
I’ve done some rounding, so be aware of that.
well divide that by 6.022x10^23 atoms (which is the amount of things in a mol)
to get how many mols of NH3
then multiply the mols by the molecular mass (14g+1x3g per mol of NH3)
to get the mass of 8.50x10^22 molecules of NH3
i got about 2.4 grams
u can do significant figures by calculating with more precise molar masses
use Avogadro's as the conversion factor.
it is the number of molecules in one gram mole = 6.022*10^23
Molecular mass of NH3 = 17g/mole which contains Avogdro's number of molecules
17*8.50x10^22/6.022*10^23 = 24*10^(22-23) = 2.4g
Avogadro's #, 6.022x10^23, is the # of atoms or molecules in 1 mole of anything.
Hence if you want to find out how many molecules is so many moles of something, you put Avogadro's # on the bottom (Denominator). If you want to find moles, you put it on the top (numerator) of the fraction.
Here's some examples:
Find how many atoms in 4.398 moles H2
H2 is a diatomic molecule, so you use molecules, then multiply by 2 to get atoms, because there are 2 atoms/molecule.
Set up: what you have, X (Avogadro's #) (atoms/molecule)
Hence: 4.398 mols H2 (6.022x10^23 molecules/1mol) = your answer, because moles cancel out.
math: 4.398* 6.022x10^23 = 2.65x10^24 molecules (2atoms/1molecule) = 5.3x10^24 atoms
Find mols 8.3x10^18 atoms Mg
8.3x10^18atoms (1mol/6.022x10^23 atoms) = mol Mg
math: 8.3x10^18/6.022x10^23 = 1.35x10^-5, or 0.0000135mol Mg
Hope this helps, good luck on your test!