Calculate the Empirical Formula for the following compound:
0.300 mol of S and 0.900 mole of O.

Explanation:

An empirical formula shows the smallest whole-number ratio of the atoms in a compound.

So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.

The first step is division. We divide the amount of moles for both elements by the smallest amount of moles.

There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300

The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.

This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g