titration question, please help (1 Viewer)

xanny

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Hey guys, i am having some trouble with this titration question, i hope someone will figure it out. thanks!

3.00 g of oxalic acid (C2H2O4) reacts with 42.0 mL of 1.586 M NaOH in a titration. Calculate how many moles of oxalic acid there are in the 3.00 g sample and the molar mass of oxalic acid.
 

honky tonk

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Hey there,

Balance the equation: C2H2O4 + 2 NaOH --> Na2C2O4 + H2O

Stoichiometry: 1 : 2 (oxalic acid : sodium hydroxide)

Moles of sodium hydroxide:
moles = molarity x volume​
moles = 1.586 x 0.042​
moles = 0.066612​

Moles of oxalic acid: Because of the 1 : 2 relationship, there are twice as many moles of sodium hydroxide than there are moles of oxalic acid, so:
moles = (0.066612 / 2)​
moles = 0.033306

Molar mass of oxalic acid:
moles = mass/molar mass​
0.033306 = 3.00/x​
x = 90.07 g
 

steamroller60

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nice work honky tonk!

i can't be bothered checking. but it looks fine.

what are you up to now? (since you say you're hsc 03)
 

Serius

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yes well done, its right :)

the formulas he used are n = cv [ moles = molarity X volume]
and n=m/MM [ moles = mass/ molar mass]

so the steps are

firstly make a balanced equation to work out the relationship e.g 2:1

then sub in values using n = cv , if relationship is 1:1 then both have the same n value, they always give enough numbers so u can work out atleast one n value

then using that n value if relationship is 2:1, divide by 2 to get the right mole

now you have in n = m/MM the n value, and they should give u in the question either the mass used, or the MolarMass used so u can work out the rest by simple manipulation of the equation
 

myeewyee

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:S Why can't you just work out the molar mass using the periodic table, and then work out the number of moles from there?
 

honky tonk

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myeewyee said:
:S Why can't you just work out the molar mass using the periodic table, and then work out the number of moles from there?
You can, but the question asked to calculate both values. It's testing your ability to understand the fundamentals of stoichiometry and the M=mol/V formula, not just the mol=mass/mw formula.
 

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