One compound contains 27.2 g of carbon and the other has 42.9 g of carbon. How can these data support the law of multiple proportions if 42.9 is not a multiple of 27.2?
2007-12-07 06:43:44 · 3 answers · asked by Sean D 2 in Science & Mathematics ➔ Chemistry
This question seems designed to confuse you; I hate that sort of thing.
All that matters are ratios of percentages, not percentages themselves. So no one except a moronic textbook writer would directly compare 27.2 with 42.9.
What IS of value is to compare the ratios (27.2 to 72.8) and (42.9 to 57.1).
Take them both down to 1 to (you work it out) and see what happens.
Working out the formulas knowing atomic masses is a bit beside the point; this is how people first DISCOVERED the idea of atomic mass.
2007-12-07 06:57:24 · answer #1 · answered by Facts Matter 7 · 1⤊ 1⤋
enable's see// If the compound is categorized like CxHyOz, then by ability of it is combustion, by ability of balancing the equation (assume that we've a million mol of CxHyOz) we get x moles of Co2 and y/2 moles of H2O. CO2........H2O 44x...........(y/2)*18 a million.727.........0.7068 by ability of fixing this we get y=2x. And now enable's revise CxHyOz: CxHyOz.....Co2 12x+y+16z....44x 0.864.... a million.727 exchange the y and sparkling up the proportion: We get z=0.5x so x:y:z=x:2x:0.5x=2:4:a million formulation is C2H4O
2016-11-13 23:55:24 · answer #2 · answered by du 4 · 0⤊ 0⤋
Yes
empirical formulas are CO2 and CO
2007-12-07 06:54:52 · answer #3 · answered by skipper 7 · 0⤊ 1⤋