anonymous
  • anonymous
Use the mass values of each element to determine the empirical formula of the tin oxide compound. tin: 2.12 g oxygen 4.5 g
Chemistry
katieb
  • katieb
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Rushwr
  • Rushwr
What did u get @JoannaBlackwelder
anonymous
  • anonymous
@iambatman can you help?
anonymous
  • anonymous

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Photon336
  • Photon336
It's been a while but I think this is how you would approach a question like this: STEP #1 convert each to moles \[2.10g Tin * \frac{ mol }{ 119.0g } = 0.18 moles \] \[ 4.5g*\frac{ mol Oxygen }{ 16g } = 0.28 \] STEP #2 find out which one has the least number of moles This is Tin 0.18 moles \[Ti _{x}O _{y}\] STEP #3 Divide each by the lowest number of moles so we're going to divide each one 0.18. 0\[\frac{ 0.18 }{ 0.18 } = 1 Tin \] \[\frac{ 0.28 }{ 0.18 } = 1.5 round \to nearest whole number 2 \] Hence Titanium di oxide. \[TiO _{2}\] let me know if you have any questions
anonymous
  • anonymous
wouldn't the empirical formula be something along the lines of SnO?
Photon336
  • Photon336
@plohrb you are correct, i accidentally put the wrong abbreviation but the molar mass of tin is still 119 g/mol though
Photon336
  • Photon336
\[SnO _{2}\]
anonymous
  • anonymous
so SnO^2 is the empirical formula or the regular?
anonymous
  • anonymous
nvm got it ! thanks!
Photon336
  • Photon336
well it would be empirical because you cant go lower than that. that's the lowest possible ratio.

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