To yield a total of 6 grams sodium citrate, how many grams of citric acid will you need?
To determine how many grams of citric acid are needed to yield a total of 6 grams of sodium citrate, we need to consider the molecular weights and stoichiometry of the two compounds.
The molecular weight of sodium citrate (Na3C6H5O7) is approximately 258.07 g/mol. From this information, we can find that the molar mass of sodium in sodium citrate is 22.99 g/mol (22.99 g/mol x 3 = 68.97 g/mol) and the molar mass of citrate is 189.06 g/mol (189.06 g/mol + 22.99 g/mol x 3 = 258.07 g/mol).
The molar ratio between sodium citrate and citric acid (C6H8O7) is 1:3. This means that one mole of sodium citrate corresponds to three moles of citric acid.
To calculate the amount of citric acid needed, we can set up a ratio based on the molar ratio:
1 mole sodium citrate : 3 moles citric acid
Since we are given the total mass of sodium citrate (6 grams), we can convert this to moles using the molar mass:
6 g / 258.07 g/mol = 0.023 mol sodium citrate
Now we can use the molar ratio to determine the amount of citric acid needed:
0.023 mol sodium citrate x (3 moles citric acid / 1 mole sodium citrate) = 0.069 mol citric acid
Finally, we can convert the moles of citric acid to grams using the molar mass of citric acid:
0.069 mol citric acid x 192.13 g/mol = 13.26 grams
Therefore, you will need approximately 13.26 grams of citric acid to yield a total of 6 grams of sodium citrate.
To calculate the amount of citric acid needed to yield 6 grams of sodium citrate, we need to understand the molecular formula of both substances and their molar masses.
The molecular formula of sodium citrate is Na3C6H5O7. To determine its molar mass, we need to look up the atomic masses of the elements from the periodic table. The atomic mass of sodium (Na) is approximately 22.99 grams/mol, carbon (C) is around 12.01 grams/mol, hydrogen (H) is approximately 1.01 grams/mol, and oxygen (O) is around 16.00 grams/mol.
So, the molar mass of sodium citrate can be calculated as:
(3 × Na) + (6 × C) + (5 × H) + (7 × O)
= (3 × 22.99) + (6 × 12.01) + (5 × 1.01) + (7 × 16.00)
≈ 258.07 grams/mol
Now, let's calculate the amount of citric acid needed. The molecular formula of citric acid is C6H8O7. Again, we find the atomic masses of each element from the periodic table.
So, the molar mass of citric acid can be calculated as:
(6 × C) + (8 × H) + (7 × O)
= (6 × 12.01) + (8 × 1.01) + (7 × 16.00)
≈ 192.13 grams/mol
Now we can set up a ratio based on the molar masses of sodium citrate and citric acid:
192.13 grams of citric acid corresponds to 258.07 grams of sodium citrate.
To find out how many grams of citric acid are needed for 6 grams of sodium citrate, we can use a simple proportion:
(6 grams of sodium citrate) / (258.07 grams of sodium citrate) = (x grams of citric acid) / (192.13 grams of citric acid)
Cross-multiplying and solving for x gives us:
x = (6 grams of sodium citrate) × (192.13 grams of citric acid) / (258.07 grams of sodium citrate)
Performing the calculation:
x = (6 × 192.13) / 258.07
≈ 4.476 grams
Therefore, you will need approximately 4.476 grams of citric acid to yield 6 grams of sodium citrate.
write the balanced equation
figure how many moles of citrate you have
the equation will tell you how many moles of acid you will get
convert that back to grams