An organic compound containing only C,H and O was subjected to combustion analysis. A sample weighing 0.4758g yielded 0.765 g CO2 and 0.313 g H2O. What is the empirical formula of the compound?

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An organic compound containing only C,H and O was subjected to combustion analysis. A sample weighing 0.4758g yielded 0.765 g CO2 and 0.313 g H2O. What is the empirical formula of the compound?

Chemistry
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well by using the combustion analysis method we can sort it out As for carbon % = mass of CO2/mass of org.comp x 12.00/44.00 x100 for hydrogen= mass of H2O/mass of org.comp x 2.016/18 x100 now %age oxygen = 100 - (%age of C + % of H) get so now simply putting the values............. we get C = 43.84% H= 7.36% so O=100-(43.84+7.36)=48.8% now moles calculation.............. simply div. %age/atomic mass(molarmass) so we get C=3.65 moles H=7.24..... O=3.05moles the last step is to obtain the atomic ratio so just div. the moles of elements by the least number of produced,,,,,,,,,,,,, (3.05 in present case) for C =3.65/3.05=1.2 H=7.24/3.05= 2.3 O=3.05/3.05=1 so the empirical formula is CH2O ??? is that rite Ma'm @Shaekitchen

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