How do you write chemical formulas? For example how so you right a chemical formula for calcium fuoride?
Calcium flouride would be CaF2
This is because ..
Calcium symbol - Ca
Flourine symbol - F
you would write CaF
BUT you must consider charges
calcium has a 2+ charge
and Flourine has a 1- charge.
The charges criss-cross and so the 2 charge goes on the F and the 1 charge goes on the Ca. Since 1 can be assumed we don't write it.
Your final answer would be CaF2
is the two a subscript
yes
A good response. It might be worth noting that the charges do criss-cross but the sign of the charge is ignored. For example, for sodium oxide, the charge on Na is +1 and the charge on oxygen is -2; therefore, the formula is Na2O (not Na-2O with the 2 being a subscript, Na2O. We DON'T write it as Na-2O. It is SO MUCH trouble, however, to write the subscripts, we just assume everyone knows Na2O is Na2O. Thanks for using Jiskha.
In the lab, 20.0 g of sodium hydroxide is added to a beaker giving it a total mass of 128.4 g. Sulfuric acid is added until all of the sodium hydroxide reacts. After heating, the total mass of the beaker and the dry compound is 142.3 g. Caculate the percent yeild.
To calculate the percent yield, you need to first determine the theoretical yield, which is the maximum amount of product that can be obtained from the given reactants. Then, you compare the theoretical yield to the actual yield, which is the amount of product obtained in the experiment. Finally, you use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
In this case, the reaction is between sodium hydroxide (NaOH) and sulfuric acid (H2SO4), and the product is unknown. However, we can assume that the reaction goes to completion and that all the sodium hydroxide reacts.
To find the theoretical yield, we need to calculate the amount of product that would be formed if all the sodium hydroxide reacted completely.
Step 1: Convert the mass of sodium hydroxide to moles.
Given: Mass of sodium hydroxide = 20.0 g
Molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.996 g/mol
Moles of NaOH = Mass / Molar mass = 20.0 g / 39.996 g/mol ≈ 0.500 mol
Step 2: Determine the stoichiometry of the reaction.
The balanced equation for the reaction between NaOH and H2SO4 is:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
This means that for every 2 moles of NaOH, 1 mole of Na2SO4 (product) is formed.
Step 3: Calculate the theoretical yield.
Since the stoichiometry of the reaction tells us that 2 moles of NaOH produce 1 mole of Na2SO4, we can infer that 0.500 mol of NaOH will produce 0.250 mol of Na2SO4.
Step 4: Convert moles of Na2SO4 to mass.
Molar mass of Na2SO4 = 2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.04 g/mol
Mass of Na2SO4 = Moles × Molar mass = 0.250 mol × 142.04 g/mol = 35.51 g
The theoretical yield of Na2SO4 is 35.51 g.
Now, we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
Given: Mass of the beaker and dry compound = 142.3 g
The actual yield is 142.3 g.
Percent Yield = (142.3 g / 35.51 g) x 100%
Percent Yield ≈ 400.96%
The percent yield is approximately 400.96%.
Please note that a percent yield greater than 100% is theoretically not possible, but it can occur due to experimental errors, impurities, or incomplete reactions.