I know this is review but i have tons of problems on percent composition and i need some refreshing. Thank you. :)
2007-04-11 10:46:45 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
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2007-04-12 12:44:07 · answer #1 · answered by SICK MY DUCK! 1 · 0⤊ 1⤋
Simply add up the molecular mass of each component to get the molecular mass of the overall compound. Then you can calulate the fraction of each omponent within the total. This is % composition.
so add up 2 atoms of Fe and 3 atoms of Oxygen.
Divide the Fe2 against the total and you have the % composition that is Iron. Now multiply with the 21.6 grams and you have the amount of Iron
2007-04-11 17:52:32 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
Mr of Fe= 55.84
Mr of O =16.00
Fe2 = 55.84 x 2 = 111.68
O3 = 16 x 3 = 48
Mr of FE2O3 = 111.68+48 = 159.68
so 111.68 grams of Fe makes up 159.68grams of Fe2O3
or 111.68/159.68 = 0.699 (ratio of Fe : Fe2O3)
so if i have 21.6grams of Fe2O3 then 0.699 x 21.6 = to the mass of Fe contributed.
0.699 x 21.6 = 15.09grams.
2007-04-11 18:02:55 · answer #3 · answered by curious 2 · 0⤊ 0⤋
the relative atomic mass of Fe2=110 03=48
the ratio is 110:48
therefore the mass of Fe = 21.6/158*110 = 15.038g
2007-04-11 17:51:58 · answer #4 · answered by King Simmy 2 · 0⤊ 1⤋
Fe2=55.8*2=111.6 amu
O3=16.0*3=48.0 amu
Total=159.6
111.6/159.6 = x g / 21.6g
15.1 g of iron in the sample
2007-04-11 17:53:48 · answer #5 · answered by ecolink 7 · 0⤊ 0⤋