The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of
<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>
This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas
<span><span>7014.04</span>=4.98≈5</span>
Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is
<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>
Percentage by mass is the mass of NaF present in 100 g of the solution.
the percentage by mass of NaF is 23.4 %
this means that in 100 g of solution, mass of NaF present is 23.4 g
the number of moles of NaF present - 1.33 mol
mass of NaF - 1.33 mol x 42 g/mol = 55.9 g
when there's 23.4 g of NaF - mass of solution is 100 g
therefore when there's 55.9 g of NaF - mass of solution is 100 / 23.4 x 55.9
= 239 g
mass of solution required is 239 g
I think its tweezers they pick up anything small