The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.
Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps
I can't actually answer this one if the empirical formula is not given. Luckily, I've found a similar problem from another website. The problem is shown in the picture attached. It shows that the empirical formula is CH₂O. Let's calculate the molar mass of the empirical formula.
Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol
Then, let's divide this to the molar mass of the molecular formula.
Molar mass of M.F/Molar mass of E.F = 180/30 = 6
Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
<em>Actual molecular formula = C₆H₁₂O₆</em>