Author Topic: Chemistry question(AAS) Help (Read 1696 times) Tweet Share

tolga · « **on:** February 13, 2010, 11:54:31 am »

The iron sample of breakfast cereal was determined using this technique. A 10.0g sample of cereal was burnt to ash in a covered crucible. The ashes were dissolved in concentrated nitric acid and diluted to 100ml with water.The concentration found was 7ug/mL for an absorbance of 0.410 in the cereal solution, when plotted the calibration graph.

a)What mass of iron was present in the ashes of te 10g of cereal?
b)What is the cocentration of iron, in mg per 100g, in the cereal?
c)If the mass of iron in 100g calculated in part b represents 56% of the recommended daily intake(RDI) of iron, what is the RDI in mg?
d)What mass of this cereal would you have to eat to obtain your RDI of iron?
Note: Can you please show working out for each part because im not very good at converting units.

Edmund · « **Reply #1 on:** February 13, 2010, 12:20:18 pm »

$1mL \rightarrow 7ug$

$1mL \rightarrow 7*10^{-6}g$

$1000mL \rightarrow 7*10^{-6}*10^3g$

$Concentration (Fe(NO_3)_3)=7*10^{-3}g/L= \frac{7*10^{-3}g}{241.9g/mol}mol/L=2.8937*10^{-5}mol/L$

$Fe(s)+3NO_3^-(aq) \rightarrow Fe(NO_3)_3(aq)$

$n(Fe(NO_3)_3)=C*V=2.8937*10^{-5}mol/L*0.100L=2.8937*10^{-6}mol$

$\therefore n(Fe)=2.8937*10^{-6}mol$

$\therefore m(Fe)=2.8937*10^{-6}mol*55.9=1.62*10^{-4}g$

Ok, I'm not sure if I'm right, I could have made a mistake somewhere. Also assumed Fe(III)

tolga · « **Reply #2 on:** February 13, 2010, 12:32:47 pm »

Part a is right with how you got 0.7mg/100ml as for part b you have to convert this answer intp mg/100g but im not sure how to do this the rest is all wrong.

Search the forums now!

We have moved!

Author Topic: Chemistry question(AAS) Help (Read 1696 times) Tweet Share

tolga

Chemistry question(AAS) Help

Edmund

Re: Chemistry question(AAS) Help

tolga

Re: Chemistry question(AAS) Help

Recent Posts

User:
Password: