anonymous
  • anonymous
Use the mass values of each element to determine the empirical formula of the tin oxide compound. mass of tin: 2.12g mass of oxygen: 4.5g
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@DanJS
DanJS
  • DanJS
sorry, i would have to look it up, been a few years since chem, it is the reduiced formula...need to find the mass ratios i think
DanJS
  • DanJS
is there more than one oxide of tin mayhbe too

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anonymous
  • anonymous
i dont think so, otherwise there would be the roman numerals right
DanJS
  • DanJS
yes, here are a couple examples, seems straight forward
DanJS
  • DanJS
https://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm
DanJS
  • DanJS
what is the formula for the tin oxide? tin has more than one oxidation state
DanJS
  • DanJS
SnO SnO2 ?
anonymous
  • anonymous
its SnO
anonymous
  • anonymous
also it says tin (II) oxide
DanJS
  • DanJS
right that is what it is if SnO , since oxygen is -2 ion charge
DanJS
  • DanJS
it just follows example 1 in that link i put above
DanJS
  • DanJS
convert both gram masses to moles of each first...
anonymous
  • anonymous
im not sure how to do that,
anonymous
  • anonymous
do i find it on the periodic table or something ?
DanJS
  • DanJS
Look on the periodic table to find the mass of each element...molecular mass
anonymous
  • anonymous
118.71 u for tin and 15.9994 u ± 0.0004 u for oxygen
DanJS
  • DanJS
that is how many grams of that element are in 1 mole of that element
DanJS
  • DanJS
sounds right, i know oxygen is 16
anonymous
  • anonymous
ok, what to do from here
DanJS
  • DanJS
so convert each to moles by.... |dw:1441161268321:dw|
DanJS
  • DanJS
notice grams X cancels out, it is on top and bottom
anonymous
  • anonymous
yea
DanJS
  • DanJS
\[\frac{ 2.12g~Sn }{ 1}*\frac{ 1~mol~Sn }{ 118.71g~Sn }~~ \approx~~0.0179~mol~~Sn\]
DanJS
  • DanJS
remember the number from the table, 118.71 gives you the number of grams of tin per 1 mole... the second fraction used
anonymous
  • anonymous
so tins molar mass is 0.0179 mol
anonymous
  • anonymous
yes
DanJS
  • DanJS
no, tins molar mass is 118.71 g / mol (on table) 0.0179 mol tin is the same thing as 2.12 g tin that is given, just converted grams of tin to moles of tin
DanJS
  • DanJS
in the same way, convert the given grams of oxygen to moles of oxygen
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
about 0.2813
anonymous
  • anonymous
is that correct
anonymous
  • anonymous
I've gotta hit the sack now but we can continue this later, thanks!
DanJS
  • DanJS
yes
DanJS
  • DanJS
follow example 1 in here https://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm already converted to moles...only one step left
DanJS
  • DanJS
goodluck
anonymous
  • anonymous
@DanJS ok im back, the last step is the division correct?
DanJS
  • DanJS
oh, yeah i think you just divide by the smallest element value
anonymous
  • anonymous
and we have to do that for each element
DanJS
  • DanJS
0.0179 mol Sn and 0.2813 mol O one sec
anonymous
  • anonymous
ok
DanJS
  • DanJS
k
DanJS
  • DanJS
yeah i think you just divide both by the smaller, and they should be near whole numbers
DanJS
  • DanJS
so 16 oxygen and one tin i guess
DanJS
  • DanJS
yeah, 4.5 grams of a gas is a bunch compared to 2.12g of tin
anonymous
  • anonymous
ok, so theres 16 oxygen and 1 tin, how does this relate to the empirical formula of tin oxide
DanJS
  • DanJS
idk, i would look more into it, i just went through it with you and looked it up
DanJS
  • DanJS
Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a compound. i'm not very confident in that answer though, it may be right, not sure
DanJS
  • DanJS
SnO16 ...looks strange
DanJS
  • DanJS
i would re-ask the question in a fresh thread and maybe someone can do it for sure
anonymous
  • anonymous
ok thanks for your time
DanJS
  • DanJS
welcome, i remember a few chem things pretty good from the 2 semesters i had to take
DanJS
  • DanJS
empirical formula, not so much, never used again

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