Question

Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, you know that phosphorus has a density of 1.82 g/cm3 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl3 (density = 1.492 g/mL) and CHBr3( = 2.890 g/mL)? (Note: 1 mL = 1 cm3.)

Answer 1

first set up the overall density equation of the solution, which is said to be equal to phosphorus' density

m_solution / v_solution = d_solution
v_solution is given as 20.0 mL
d_solution is given as 1.82 g/cm3 or 1.82g/mL
you can then solve for the total mass of the solution

let's call the mass of CHCl3 m1 and the mass of CHBr3 m2

m1 + m2 = m_solution

now, we can re-write individual density equations for the Cl and Br masses as

d1 = m1/v1
d2 = m2/v2

or, solving for m1 and m2

m1 = d1 * v1
m2 = d2* v2

Answer 2

why do this? m1 + m2 = m_solution is not enough to solve for the unknowns m1 and m2
but if we re-write in terms of density and volume, we can also use the fact that v1 + v2 = 20mL to set up a system of equations basically

so, substituting these new equations into the m1 + m2 = m_solution equation

d1 * v1 + d2 * v2 = m_solution

you can let v1 = 20 - v2. solve the equation for v2, then back-substitute to find v1.