Question

mass: 3.00 g
h₂o
so₂
ru(co)₅
sf₆
cf₄
xef₄
molar mass
18.015
64.058
241.12
146.05
88.004
207.29
percent compositions
11.18 h 88.81 o
50.04 s 49.96 o
41.91 ru 24.90 c 33.17 o
21.95 s 78.04 f
13.64 c 86.35 f
63.34 xe 36.35 f
mass (g)\tamount (mol)\tparticles
h₂o\t3.00\t0.167\t1.00 × 10²³
h\t0.335\t0.333\t2.01 × 10²³
o\t2.66\t0.167\t1.00 × 10²³
now examine the mass of each element in an sf₆ sample. set the mass of sf₆ to 5.4 g.
what is the ratio
\frac{mass f atoms}{mass s atoms}?

Explanation:

Step1: Find the percent composition of F and S in \(SF_6\)

From the given data, for \(SF_6\), the percent composition of \(S = 21.95\%\) and \(F=78.04\%\)

Step2: Calculate the ratio \(\frac{\text{mass F atoms}}{\text{mass S atoms}}\)

Let the mass of the \(SF_6\) sample be \(m\). The mass of \(F\) atoms in the sample, \(m_F=0.7804m\), and the mass of \(S\) atoms in the sample, \(m_S = 0.2195m\)
The ratio \(\frac{m_F}{m_S}=\frac{0.7804m}{0.2195m}\)

Since \(m
eq0\) (because we have a non - zero mass sample of \(SF_6\)), we can cancel out \(m\)

\(\frac{m_F}{m_S}=\frac{0.7804}{0.2195}\approx3.555\)

Answer:

\(3.56\) (rounded to three significant figures)