The formula of tartaric acid is is H6C4O6 and it is a dibasic acid?

Real Chemist

Verified answer

(i)

H6C4O6 + NaOH ----> NaH5C4O6 + H2O

NaH5C4O6 __ or __ NaOOC-CH(OH)-CH(OH)-COOH

Sodium hydrogentartrate

H6C4O6 + 2 NaOH ----> Na2H4C4OO6 + 2 H2O

Na2H4C4O6 __ or __ NaOOC-CH(OH)-CH(OH)-COONa

di-Sodium tartrate

(iI)

Copper (II) tartrate

CuH4C4O6

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There are a number of methods used to do this calculation. I have settled on the one I use below because I think that it is easy to follow and it reinforces the mol concept which I encourage my students use and understand. First a little theory which will make the understanding easier. The original tartaric acid you know is a compound consisting only of C, H and O. The substance burns or is oxidised by oxygen to produce CO2 and H2O All the C in the original compound is converted to CO2 ( do not be concerned where the O comes from - it is unimportant) You know the mass of CO2 produced - it is given to you - so ytou can calculate the mass of C originally in the 12.01g acid sample Likewise with the H. All the H in the original acid was oxidised to water. You know the mass of water produced so you can calculate the mass of H in the sample If you know the mass of C and H, by difference yiu can calculate the mass of O in the 12.01g sample. Let us do these steps: Mass C in 14.08g CO2 Molar mass C = 12.011g/mol Molar mass CO2 + 44.0096g/mol Mass C = 12.011/44.0096*14.08 = 3.84268g Mass H in 4.32g H2O Molar mass H = 1.008: 2H = 2.016g/mol Molar mass H2O = 18.0015g/mol Mass H = 2.016/18.0015*4.32 = 0.4838g Mass O = 12.01 - ( 3.84268 + 0.4838) = 7.6835g These mass values have to be brougfht into moles of each element. Divide by the respective atomic mass: C = 3.84268/12.001 = 0.3199 H = 0.4838 / 1.008 = 0.4799 O = 7.6835/ 15.999 = 0.4802 To get these mol values into a ratio, divide by the smallest C = 0.3199/0.3199 = 1 H = 0.4799/0.3199 = 1.5 O = 0.4802/0.3199 = 1.5 You cannot write CH1.5O1.5 - you have to have whole integers, so multiply through by 2 C = 2 H = 3 O = 3 Empirical formula = C2H3O3

Anónimo

H2O is 11.1% H and 88.8% O, so 11.1% of 4.32g = 0.48g H and 3.84g O CO2 is 27.3% C and 72.7%O, so 27.3% of 14.08g = 3.84gC and 10.24g O find moles of each element 0.48gH / 1g/mole = 0.48 moles H 3.84gC / 12g/mole = 0.32 moles C 7.69gO / 16g/mole = 0.48 moles O to determine the grams O present, we have to add up the O found in CO2 + the O found in H2O. this gives us 14.08g O. this is more than the tartaric acid alone so, as we know, the extra oxygen came from the O2 that was used to oxidize the tartaric acid. the total mass of the tartaric acid is 12.01g, 0.48 of which is H, 3.84g of which is C, the rest is Oxygen, or 7.69g divide by the smallest number to get the ratios 0.48H / 0.32 = 1.5 H 0.32O / 0.32 = 1 C 0.48C / 0.32 = 1.5 O the formula is C1 H1.5 O1.5 since we cannot have fractions of moles in a formula, determine the common factor by which to multiply everything to get rid of the fractions, in this case, 2 ***C2H3O3 is the empirical formula