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Chemistry 11 Online
OpenStudy (haileyd):

Finding percent composition..

OpenStudy (haileyd):

OpenStudy (haileyd):

@matt101

OpenStudy (photon336):

@HaileyD do you know how to approach this?

OpenStudy (haileyd):

So when we did a problem we put the number of mol times molar mass is that right cause i tried that and it didnt seam correct.

OpenStudy (photon336):

so you know the total mass of this compound right? and the percentage of each element. you can easily find the number of grams of each element present in that compound. we can do this by multiplying the total number of grams 180 by the percentage of each element. see below. \[180grams*(\frac{ 6.67 }{ 100 }) = 12.0~grams~Hydrogen \] \[180*(\frac{ 53.3 }{100 }) = 95.94~grams~oxygen\] \[180*(\frac{ 40 }{ 100 }) = 72~grams~Carbon\]

OpenStudy (haileyd):

Okay thank you

OpenStudy (photon336):

we're not done yet, we need to find the formula for the compound

OpenStudy (photon336):

so now you know how many grams of each compound we have, the next thing to do is convert each to moles. \[12.0grams~x \frac{ mol }{ 1gram } = 12~moles~of~Hydrogen\] \[95.94~grams*(\frac{ mol }{ 16 grams}) = 6.0~mol~of~Oxygen\] \[72~grams~carbon*(\frac{ mol }{ 12grams }) =6 ~moles of~carbon \] Molecular formula \[C_{6}H_{12}O_{6}~Glucose \] Empirical formula now, if we wanted to find the empirical formula for the compound that is the lowest ratio of atoms in the compound we could divide each # of moles but the smallest number in this case 6 Hydrogen 12/6 =1 Carbon 6/6 = 1 Oxygen 6/6 = 1 \[CH_{2}O\]

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