Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.
First, let's find the total mass by using the masses of the elements given on the periodic table.
7 x 12.011 (mass of Carbon) = 84.077
5 x 1.008 (mass of Hydrogen) = 5.04
3 x 14.007 (mass of Nitrogen) = 42.021
6 x 15.999 (mass of Oxygen) = 95.994
Add all of those pieces together.
84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.
Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %
Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %
Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %
Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %
You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.
Answer:
No. 4
Explanation:
Bacteria are produced from pre existing bacteria
First, you need to be given the gram formula mass of this molecular or related conditions that can calculate this. Then you need the empirical formula and calculate the gram formula mass of empirical formula. Then use them to get the ratio. Finally you can get the molecular formula.
Answer:
Concentration solution A was 0.5225 M
Explanation:
10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B
According to laws of dilution- ![C_{A}V_{A}=C_{B}V_{B}](https://tex.z-dn.net/?f=C_%7BA%7DV_%7BA%7D%3DC_%7BB%7DV_%7BB%7D)
where,
and
are concentration of solution A and B respectively
and
are volumes of solution A and B respectively
Here
= 0.1045 M,
= 50.00 mL and
= 10.00 mL
Hence, ![C_{A}=\frac{C_{B}V_{B}}{V_{A}}=\frac{(0.1045M\times 50.00mL)}{10.00mL}=0.5225M](https://tex.z-dn.net/?f=C_%7BA%7D%3D%5Cfrac%7BC_%7BB%7DV_%7BB%7D%7D%7BV_%7BA%7D%7D%3D%5Cfrac%7B%280.1045M%5Ctimes%2050.00mL%29%7D%7B10.00mL%7D%3D0.5225M)
So, concentration solution A was 0.5225 M
Answer:
The mole fraction of CH3OH in this solution is closest to 0.1.
Explanation:
The mole fraction of CH3OH can be calculated by dividing the number of mole of CH3OH by the sum of all moles present in the solution. In our example, we have 16 g CH3OH and 90 g H2O. Let´s see how many mols of each constituent we have:
1 mol CH3OH = 32 g ⇒ 16 g CH3OH = 0.5 mol.
1 mol H2O = 18 g ⇒ 90 g H20 = 5 mol
Then, the mole fraction is:
X = mol CH3OH / (mol H2O + mol CH3OH)
X= 0.5 mol / (5 mol + 0.5 mol) ≅ 0.1