Sublimation of Arsenic(III) sulfide

 

Arsenic(III) sulfide sublimes readily, even below its melting point of 320 °C. The molecules of the vapor phase are found to effuse through a tiny hole at 0.28 times the rate of effusion of Ar atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

This should essentially be an effusion-rate problem. I know the equation is [math]\dfrac{r_x}{r_y}=\sqrt{\dfrac{M_y}{M_x}}[/math], where [math]r[/math] is the rate of effusion and [math]M[/math] is molar mass.

I know [math]M_{Ar}=39.95[/math] g and the effusion rate for arsenic(III) sulfide is [math]r=.28r_{Ar}[/math].

Therefore, [math]\dfrac{r_{Ar}}{0.28r_{Ar}}=\sqrt{\dfrac{M}{39.95g}}[/math]. I'm trying to find [math]M[/math].

After the math, I get [math]M=509.6[/math] g (to 4 significant figures).

The closest working ratio for the arsenic compound seems to be two formula units, As₄S₆. This is the textbook answer. However, this molar mass is [math]492.04[/math] g, a bit far off from my answer. Why? Did I go about this wrong?