Question #bd620

Notice that the chemical formula of sodium chlorate tells you that #1# mole of this compound contains

• #1# mole of sodium atoms, #1 xx "Na"#
• #1# mole of chlorine atoms, #1 xx "Cl"#
• #3# moles of oxygen atoms, #3 xx "O"#

This means that if you take a sample of sodium chlorate that contains exactly #1# mole of sodium chlorate, you can use the molar masses of sodium chlorate and of its constituent elements to say that you have

#overbrace("106.4412 g")^(color(blue)("1 mole of NaClO"_3)) = overbrace("22.99 g")^(color(blue)("1 mole of Na")) + overbrace("35.453 g")^(color(blue)("1 mole of Cl")) + overbrace(3 xx "15.9994 g")^(color(blue)("3 moles of O"))#

This tells you that for every #"106.4412 g"# of sodium chlorate, the equivalent of #1# mole of this compound, you get #"35.453 g"# of chlorine atoms, the equivalent of #1# mole of chlorine.

Therefore, you can say that your sample will have a mass of

#142 color(red)(cancel(color(black)("g Cl"))) * "106.4412 g NaClO"_3/(35.453color(red)(cancel(color(black)("g Cl")))) = color(darkgreen)(ul(color(black)("426 g NaClO"_3)))#

The answer is rounded to three sig figs.