Determine the molar mass of Mn (ClO4)3. Caroline knows the amount of a compound

Question

Determine the molar mass of Mn (ClO4)3. Caroline knows the amount of a compound in moles. To find the number of molecules in the given sample of her compound, which of the following must also be used in her calculation? Avogadro's number atomic mass of each element in the compound molecular mass of the compound mass of the sample in grams Calculate the following in 9.71 g U3 PO4 What is the formula for ammonium sulfide? Capitalization and punctuation count! How many hydrogen atoms are in 5.80 mol of ammonium sulfide?

Explanation / Answer

1) Molar mass of Mn(ClO4)3 = 353.29 g/mole

2) Molar mass of Li3PO4 = 115.79 g/mole

Thus, moles of Li3PO4 in 9.71 g of it = mass/molar mass = 9.71/115.79 = 0.084

Thus, moles of Li+ = 3*moles of Li3PO4 = 0.252

Thus, number of Li+ ions = moles*Avagadros' number = 1.518*1023

Moles of O = 4*moles of Li3PO4 = 0.336

Thus, mass of O = mole*molar mass = 0.336*16 = 5.376 g

3) Ammonium sulfide = (NH4)2S

moles of H-atoms in 5.8 moles of ammonium sulfide = 5.8*8 = 46.4 moles

Thus, number of H-atoms = moles*Avagadro's Number = 2.79*1025 atoms

4) Number of molecules = moles*Avagadro's number

Thus, the correct option is :- (A)

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