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Report (any type)/Brief report
22/09/2021
##”Fe”_3″O”_4##
So, if you assume that all the mass of iron took part in the reaction, you can say that the difference between the mass of the product and the mass of iron will represent the mass of oxygen.
##m_”product” = m_”iron” + m_”oxygen”##
##m_”oxygen” = “118.37 g” – “85.65 g” = “32.72 g”##
Now that you know how much iron and how much oxygen you have in the iron oxide, you can use the two ‘ respective molar masses to figure out how many moles of each you have.
For iron, you will have
##85.65color(red)(cancel(color(black)(“g”))) * “1 mole Fe”/(55.845color(red)(cancel(color(black)(“g”)))) = “1.534 moles Fe”##
For oxygen, you will have
##32.72color(red)(cancel(color(black)(“g”))) * “1 mole O”/(16.0color(red)(cancel(color(black)(“g”)))) = “2.045 moles O”##
To determine the empirical formula of the iron oxide, you need to find the smallest whole number ratio that exists between the two elements in the compound.
To do that ,divide both values by the smallest one.
##”For Fe: ” (1.534color(red)(cancel(color(black)(“moles”))))/(1.534color(red)(cancel(color(black)(“moles”)))) = 1##
##”For O: ” (2.045color(red)(cancel(color(black)(“moles”))))/(1.534color(red)(cancel(color(black)(“moles”)))) = 1.333##
To get the smallest whole number ratio that exists between the two elements, multiply both values by ##color(blue)(3)##. This will help you get rid of the decimal fraction for oxygen.
The empirical formula of the iron oxide will thus be
##(“Fe”_1″O”_1.333)_color(blue)(3) implies color(green)(“Fe”_3″O”_4)##
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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