2) It takes about 0.69 µL (microliters) of a copper solution containing 0.4403 g

ID: 892023 • Letter: 2

Question

2) It takes about 0.69 µL (microliters) of a copper solution containing 0.4403 g of copper ion per 100 mL to make a spot approximately 0.494 cm in diameter.
Calculate the micrograms of copper ion in the spot.

Using the "ruler" in the picture of the chromatogram and measuring with an accuracy of .1 unit,
calculate the retardation factor (Rf) for compound...... A.

MEASUREMENTS AND CALCULATIONS

Why do 1a, 1b, and 2 have units wheras 1c does not?

1a) How far did compund A move? ______________ "units". 1b) At the same time, how far did the solvent move? ______________ "units". 1c) What is the Rf for compound A? ______________ . 2) How many µg (micrograms) of copper ion are in the spot? ______________ µg. 8 Solvent Front line 7 6 5 4 2 1 Spotting line 0

Explanation / Answer

0.4403 g/ 100 mL = 4.403 g / L.

Now change the unit from g to ug and L to uL as follows:

(4.403 g/L) x (1.000 e+6 ug/g) x (1.000 L / 1.000 e+6 uL) = 4.403 ug/uL

Now Calculate the micrograms of copper ion in the spot

4.403 ug/uL x (0.69 uL) = 3.03ug of copper ion.

MEASUREMENTS AND CALCULATIONS

1a) How far did compund A move?

2 cm approximatly "units".

1b) At the same time, how far did the solvent move?

6.2 cm approximatly "units".

1c) What is the Rf for compound A?

0.32.

2) How many µg (micrograms) of copper ion are in the spot?

_= 3.03ug of copper ion µg.

Why do 1a, 1b, and 2 have units wheras 1c does not?

Because Rf is the ratio of two unit that why it is unit less.

Rf = distance moved by the compound or spot / distance moved by the solvent.

Rf = 2cm / 6.2 cm= 0.32