1. Calculate the molar mass of Ca3(PO4)2. 2. How many molecules are contained in
ID: 1001959 • Letter: 1
Question
1. Calculate the molar mass of Ca3(PO4)2. 2. How many molecules are contained in the 254 grams of CH4? The molar mass of CH4 is 16 grams/mole. 3. What is the mass of 6.89*1022 molecules of CO? The molar mass of CO is 30 grams/ moles. 4. Calculate the mass percent composition of potassium in K3PO4. 5. Determine the molecular formula of a compound that has a molar mass of 184 g/mol and an empirical formulas of NO2. 6. Determine the empirical formula of a compound that has 16.51 grams of Ag and 1.224 grams of O by mass 7. write the balanced equation to show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite.Explanation / Answer
1) Ca3(PO4)2 MW = sum of MW of all atoms present in the sustance so
Ca = 40g/mol x 3 = 120g/mol
P = 31g/mol x 2 = 62g/mol
O = 16g/mol x 8 = 128g/mol
MW of Ca3(PO4)2 MW =(120 + 62 + 128 )g/mol = 310g/mol
2) 1 mol have 6.02x1023molecules so m/MW = n number of moles of CH4 and MW of CH4 = 16g/mol
254g / 16g/mol = 15.9mol as 15.9mol x 6.02x1023molecules/mol = 9.6x1024molecules
3) 6.89x1022molecules of CO2 as 1 mol have 6.02x1023molecules so
6.89x1022molecules/6.02x1023molecules/mol = 0.12mol as the MW of CO2 = 44g/mol
m = n x MW m = 0.12mol x 44g/mol = 5.28g
4) so equal in 1
MW K = 39.1g/mol x 3 = 117.3g/mol and MW PO4 = 95g/mol MW K3PO4 = (117.3 + 95)g/mol = 212.3g/mol
the % of K in K3PO4 =100% MW K x 3 / MW K3PO4 = 100% 117.3g/mol / 212.3g/mol = 55.3%
5) MW N = 14g/mol MW O = 16 MW NO2 = (14 + 16x2 ) g/mol = 46g/mol as MW of sustance is 184
the relation is 184g/46g/mol =4 so the sustace have a formula 4NO2 = N4O8
6) MW Ag = 108g/mol MW O = 16g/mol n Ag = 16.51g/108g/mol = 0.15mol and n O = 1.224g/16g/mol = 0.08mol the relation of Ag/O = to your composition molecular = 0.15mol/0.08mol = 2 so we have 2mol of Ag per mol of O so the molecular formula is Ag2O
7) 2H2SO4 + 4LiOH -------> 2Li2SO3 + 4H2O + O2
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