Question

Determination of the relative molecular mass of compound (a), any help ???

We dissolved an unknown monobasic organic acid A in exactly `"50.0 cm"^3` of 0.93 mol/L `"NaOH"`. Then we back-titrated with 0.1000 mol/L sulfamic acid.

So I did my titration and we were given a table:

Mass of compound A:
Sample 1: 0.3398 g
Sample 2: 0.3322 g

Volume of sulfamic acid solution:
Sample 1: `"35.36 cm"^3`
Sample 2: `"32.90 cm"^3`

Need to work out for both samples:
Moles of NaOH solution
Moles of compound A
`"Molecular Weight of A/g·mol"^"-1"`
Average Molecular weight of A
And the Melting point of compound A is 122-177 °C
From that and molecular weight I need to identify compound A.

Answer 1

Warning! Long Answer. Here's what I get.

Explanation:

1. Moles of `"NaOH"`

`"Moles of NaOH" = "0.0500 dm"^3color(white)(l) "NaOH" × "0.093 mol NaOH"/(1 "dm"^3color(white)(l) "NaOH") = color(red)(4.65 × 10^"-3"color(white)(l) "mol NaOH")`

2, Moles of compound A

(a) Concentration of sulfamic acid

`ulbb("Sample"color(white)(mmm)1color(white)(mmmmml)2color(white)(mmmm))`
`m"/g":color(white)(mmml)0.3398color(white)(mmmll)0.3322`
`n"/mol":color(white)(mml)3.500 ×10^"-3"color(white)(m)3.422 × 10^"-3"`
`V"/dm"^3:color(white)(mll)50.0color(white)(mmmmll)50.0`
`c"/mol·L"^"-1":color(white)(m)0.1750color(white)(mmmll)0.1711`

(b) Moles of sulfamic acid used in back-titration

`ulbb("Sample"color(white)(mmm)1color(white)(mmmmmm)2color(white)(mmmm))`
`V"/dm"^3:color(white)(mml)35.36color(white)(mmmml)32.90`
`c"/mol·dm"^3":color(white)(m)0.1000color(white)(mmml)0.1000`
`n"/mol":color(white)(mmm)3.536 ×10^"-3"color(white)(m)3.290 × 10^"-3"`

(c) Excess moles of `"NaOH"`

The equation for the reaction is

`"NH"_2"SO"_3"H" + "NaOH" → "NH"_2"SO"_3"Na" + "H"_2"O"`

`ulbb("Sample"color(white)(mmmmmmll)1color(white)(mmmmml)2color(white)(mmmm))`
`"Moles of SA":color(white)(mmll)3.536 ×10^"-3"color(white)(m)3.290 × 10^"-3"`
`"Moles of NaOH":color(white)(m)3.536 ×10^"-3"color(white)(m)3.290 × 10^"-3"`

(d) Moles of NaOH neutralized by A

Moles neutralized by A = original moles - excess moles

`ulbb("Sample"color(white)(mmmmmmll)1color(white)(mmmmml)2color(white)(mmmm))`
`"Original moles":color(white)(ml)4.65 ×10^"-3"color(white)(ml)4.65 × 10^"-3"`
`"Excess moles":color(white)(mm)3.536 ×10^"-3"color(white)(m)3.290 × 10^"-3"`
`"Moles by A":color(white)(mmm)1.114 ×10^"-3"color(white)(ll)1.360 × 10^"-3"`

(e) Moles of `"A"` neutralized by `"NaOH"`

The equation for the reaction is

`"HA + NaOH" → "NaA" + "H"_2"O"`

`ulbb("Sample"color(white)(mmmmmll)1color(white)(mmmmml)2color(white)(mmmm))`
`"Moles NaOH":color(white)(m)1.114 ×10^"-3"color(white)(ml)1.360 × 10^"-3"`
`"Moles A":color(red)(color(white)(mmml)1.11 ×10^"-3"color(white)(mm)1.36 × 10^"-3")`

3. Molar mass of A

`"Molar mass" = "mass"/"moles" = m/n`

`ulbb("Sample"color(white)(mmmmm)1color(white)(mmmmml)2color(white)(mmm))`
`m"/g":color(white)(mmmmml)0.3398"color(white)(mmmml)0.3350`
`n"/mol":color(white)(mmml)1.11 ×10^"-3"color(white)(mm)1.36 × 10^"-3"`
`MM"/g·mol"^"-1":color(red)(color(white)(mm)305color(white)(mmmmmll)246")`

It looks like the molar mass of A is about `color(red)("280 g")`.