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anonymous

  • one year ago

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

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  1. Rushwr
    • one year ago
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    What did u get @JoannaBlackwelder

  2. anonymous
    • one year ago
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    @iambatman can you help?

  3. anonymous
    • one year ago
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    @freckles @undeadknight26 @zzr0ck3r

  4. Photon336
    • one year ago
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    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

  5. anonymous
    • one year ago
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    wouldn't the empirical formula be something along the lines of SnO?

  6. Photon336
    • one year ago
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    @plohrb you are correct, i accidentally put the wrong abbreviation but the molar mass of tin is still 119 g/mol though

  7. Photon336
    • one year ago
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    \[SnO _{2}\]

  8. anonymous
    • one year ago
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    so SnO^2 is the empirical formula or the regular?

  9. anonymous
    • one year ago
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    nvm got it ! thanks!

  10. Photon336
    • one year ago
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    well it would be empirical because you cant go lower than that. that's the lowest possible ratio.

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