Baking Soda Chemistry Question?
This is a question from my grade 11 Chemistry book, and we have to answer this question for an assignment due tomorrow. I know he said you had to do the percentage composition of NaHCO3. I tried to do that, but I got stuck. Can someone please please please please help me???
Answer:
You would use dimentional analysis. Below is all one equation
0.5gNaHCO3 x 1moleNaHCO3 x
.......84.0069g NaHCO3..
...1mole Na .x 22.9898gNa = 0.137g
1moleNaHCO3..1mole Na
0.137g of Sodium in 0.5 grams of NaHCO3 or 137mg of Na
Sodium = 23
Hydrogen = 1
Carbon = 12
Three oxygens = 48
Total 84 g / mol
23/84*100 = 27% sodium by weight in sodium bicarbonate
0.5 x .27 = 0.137
Looking good.
Na = 22.99 g/mol
NaHCO3 = 84.007 g/mol
so if there is 0.500g of NaHCO3, there should be 23/84 * 0.500 or 0.1369g, or 136.9mg.
Hence, it is correct.
Step 1: Find the molecular mass of NaHCO3.
Mr = 23 + 1 + 12 + (16 x 3) = 84g/mol.
Step 2: Find the mass of sodium in the molecule.
Periodic table says 23.
Step 3: Find the percentage of the molecule that is sodium:
(23/84) x 100 = 27.3%
So in any given sample of NaHCO3, 27.3% of it must be sodium.
Step 4: Find the percentage composition in 0.500g:
0.500 x 27.3% = 0.137g of sodium = 137mg of sodium.
Seems the box doesn't lie...
MW NaHCO3 = 22.99+1.008+12.011+16x3=84.01
1 mole of NaHCO3 weighs 84.01 gram
1 mole of NaHCO3 contains 1 mole of Na atoms or 22.99 grams
The ratio of the weight of Na to NaHCO3 is 22.99/84.01 or 0.27366.
500 mg of NaHCO3 x 0.27366 = 136.8 mg Na
NaHCO3 ...Mol.mass = 23Na + 1H + 12C + 48(O)
= 84g/mol.
Moles of Na = 23 ÷ 84 = 0.274moles (x 100) = 27.4%
27.4% of 0.500g = 0.274 x 0.500 = 0.137g = 137mg Na.
NaHCO3 = 23+1+12+16X3 = 84
So in 500mg contains 500/84X23 = 136.90476 say 137
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