51.2% water in espon salt
1. Find the molar mass (MM) of the substances.MM FeSO4 = 151.91g+MM 7H2O = 126.11gMM FeSO4 x 7H2O = 278.02g2. Find the percent water of hydrate.Divide the mass of water by the mass of the hydrate; multiply result by 100%.126.11g278.02g x 100%Percent water of FeSO4 * 7H2O is 45.36%.Finding Molar Mass# atoms Element A x atomic mass Element A = Mass A# atoms Element B x atomic mass Element B = Mass B... etc.Add up all the mass values for the substance and you have the molar mass of the substance.
1. Find the molar mass (MM) of the hydrate (cobalt acetate tetrahydrate).MM Co(C2H3O2)2: 177.02g (anhydrate)+ MM 4H2O: 72.06g (water)MM Co(C2H3O2)2 * 4H2O = 249.08g (hydrate)2. Find the percent water of hydrate. Divide the mass of water by the mass of hydrate (anhydrate + water = hydrate).72.06g249.08g x 100%Percent water of hydrate is 28.93%.Finding Molar Mass# atoms Element A * Atomic Mass Element A = Mass A# atoms Element B * Atomic Mass Element B = Mass B... etc.Add up all of the mass values and you have the molar mass of the substance.
The percent, by weight of 17g of sucrose in 188g of water is calculated as follows Mass of Sucrose / Total Mass of Solution *100% = 17 / (188 + 17) * 100 = 8.29%
mass percent=g solute/ g solute=g solvent X 100% mass percent = 50g/ 50g+1000g X100%= 4.76%
4.2 grams NaNO3/60 grams water * 100 = 7% by mass -------------------
70% water and 30%mass
0.023%
Around 20.91% of gypsum is water by mass.
0,98
60 percent of your body is water.
1. Find the molar mass (MM) of the substances.MM FeSO4 = 151.91g+MM 7H2O = 126.11gMM FeSO4 x 7H2O = 278.02g2. Find the percent water of hydrate.Divide the mass of water by the mass of the hydrate; multiply result by 100%.126.11g278.02g x 100%Percent water of FeSO4 * 7H2O is 45.36%.Finding Molar Mass# atoms Element A x atomic mass Element A = Mass A# atoms Element B x atomic mass Element B = Mass B... etc.Add up all the mass values for the substance and you have the molar mass of the substance.
88.5%
2 out of 18 or 11.1111 %
water
Mass % of CHCl3 = 15x10-4 %
1. Find the molar mass (MM) of the hydrate (cobalt acetate tetrahydrate).MM Co(C2H3O2)2: 177.02g (anhydrate)+ MM 4H2O: 72.06g (water)MM Co(C2H3O2)2 * 4H2O = 249.08g (hydrate)2. Find the percent water of hydrate. Divide the mass of water by the mass of hydrate (anhydrate + water = hydrate).72.06g249.08g x 100%Percent water of hydrate is 28.93%.Finding Molar Mass# atoms Element A * Atomic Mass Element A = Mass A# atoms Element B * Atomic Mass Element B = Mass B... etc.Add up all of the mass values and you have the molar mass of the substance.
The percent, by weight of 17g of sucrose in 188g of water is calculated as follows Mass of Sucrose / Total Mass of Solution *100% = 17 / (188 + 17) * 100 = 8.29%