A 12.0g sample of carbohydrate contain 4.50g of carbon, 3.25g of oxygen, and some hydrogen. Determine the percent by mass of hydrogen in your product.
Find the empirical formula for a compound that consists of aluminum and chlorine in which the aluminum is 20.2 % by mass.
. A sample contains 71.65% CL, 24.27% C and 4.07% H. the molecular weight is known to be 98.96 g/mol. What are the empirical and molecular formulas?
I worked 1 and 2 for you a couple of days ago.
g H(atoms that is) = 12.0-gO - gC = ?
%H = (g H/12.O)*100 = ?
20.2% Al
100-20.2 = %Cl
Take a 100 g sample which will give you
20.2g Al
79.8 g Cl
Convert grams t mols. mol = grams/molar mass.
Then find the ratio of Al and Cl to each other with the smallest number being 1.00. That will give you the empirical formula.
#3. This done the same way as #2.
Post your work if you get stuck.
To determine the percent by mass of hydrogen in a carbohydrate sample, we need to first calculate the mass of hydrogen in the sample.
Given:
Mass of carbon = 4.50 g
Mass of oxygen = 3.25 g
To find the mass of hydrogen, we can subtract the sum of the masses of carbon and oxygen from the total mass of the sample.
Total mass of the sample = 12.0 g
Mass of hydrogen = Total mass of the sample - (Mass of carbon + Mass of oxygen)
= 12.0 g - (4.50 g + 3.25 g)
= 4.25 g
Now, we can calculate the percent by mass of hydrogen by dividing the mass of hydrogen by the total mass of the sample and multiplying by 100.
Percent by mass of hydrogen = (Mass of hydrogen / Total mass of the sample) * 100
= (4.25 g / 12.0 g) * 100
= 35.42%
Therefore, the percent by mass of hydrogen in the carbohydrate sample is 35.42%.
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To find the empirical formula for a compound that consists of aluminum and chlorine, where aluminum is 20.2% by mass, we need to calculate the moles of aluminum and moles of chlorine.
Given:
Percent by mass of aluminum = 20.2%
Percent by mass of chlorine = 100% - 20.2% = 79.8%
To calculate the moles of aluminum and chlorine, we need the molar masses of each element.
Molar mass of aluminum = 26.98 g/mol
Molar mass of chlorine = 35.45 g/mol
We can now calculate the moles of each element by dividing the mass percent by the molar mass.
Moles of aluminum = (Percent by mass of aluminum / 100) * (Total mass of the compound / Molar mass of aluminum)
= (20.2 / 100) * (Total mass / 26.98 g/mol)
Moles of chlorine = (Percent by mass of chlorine / 100) * (Total mass of the compound / Molar mass of chlorine)
= (79.8 / 100) * (Total mass / 35.45 g/mol)
Once we have the moles of aluminum and chlorine, we can find the simplest whole-number ratio by dividing each by the smallest value.
Empirical formula = (Moles of aluminum / Smallest value) : (Moles of chlorine / Smallest value)
This will give us the empirical formula.
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To determine the empirical and molecular formulas for a compound given the mass percentages and molecular weight, we can follow these steps:
Given:
Percent of CL = 71.65%
Percent of C = 24.27%
Percent of H = 4.07%
Molecular weight = 98.96 g/mol
First, convert the percentages into grams by assuming a 100 g sample. Then, calculate the number of moles for each element.
Mass of CL = (Percent of CL / 100) * 100 g = 71.65 g
Mass of C = (Percent of C / 100) * 100 g = 24.27 g
Mass of H = (Percent of H / 100) * 100 g = 4.07 g
Moles of CL = Mass of CL / Molar mass of CL
Moles of C = Mass of C / Molar mass of C
Moles of H = Mass of H / Molar mass of H
Next, determine the ratios of the elements in terms of moles. Divide each mole value by the lowest value to get the simplest whole-number ratio.
Empirical formula = (Moles of CL / Smallest value) : (Moles of C / Smallest value) : (Moles of H / Smallest value)
Now, to find the molecular formula, we need to determine the molecular weight of the empirical formula.
To calculate the molecular weight of the empirical formula, you need to sum up the atomic weights of all the atoms present in the empirical formula.
Finally, divide the given molecular weight by the empirical formula weight to find the multiple needed to get the molecular formula.
Molecular formula = Empirical formula * (Molecular weight / Empirical formula weight)
Note: Ensure that all the values are rounded to the nearest whole number for the simplest ratio and molecular formula.